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April 20, 2014

Search: N2H4(g) + H2(g) 2 NH3(g)

Number of results: 5,610

Enthalpy of Formation
N2H4(g) + H2(g)--> 2 NH3(g) H1 = –1876kJ 3 H2(g) + N2(g)--> 2 NH3(g) H2 = –922 kJ The H.f for the formation of hydrazine: 2 H2(g) + N2(g)--> N2H4(g) will be______kj/mol I am very confused here. I thought I would just add H1 and H2 and divide by the molar mass of N2H4...
Sunday, May 16, 2010 at 12:46am by CHem

Chemistry
Please :) N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be_____kJ/mol.
Friday, February 18, 2011 at 12:36am by HELP!

Chemistry
Please :) N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be_____kJ/mol.
Friday, February 18, 2011 at 12:36am by HELP!

Chemistry
N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be
Friday, February 22, 2008 at 1:43pm by Charlie

Chemistry
N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be ______ kJ/mol
Wednesday, May 13, 2009 at 2:36pm by Tylor

Chemistry
N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be _____ kJ/mol.
Saturday, February 20, 2010 at 2:02am by Cuukoo

Chemistry
N2H4(g) + H2(g) ---> 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) ---> 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be _________ kJ/ mol.
Sunday, February 24, 2008 at 3:40pm by Anonymous

chemistry
N2H4(g) + H2(g) 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) 2 NH3(g) H2 = –922 kJ The H f for the formation of hydrazine: 2 H2(g) + N2(g) N2H4(g) will be ? kJ/mol. Turn equation 1 around and change the sign of H1. Add to equation 2, cancel like molecules appearing on both sides, ...
Friday, February 9, 2007 at 11:30pm by nick

Enthalpy of Formation
You have to reverse the direction of one reaction and change the sign of H. 3 H2+ N2--> 2 NH3 H2 = -922 kJ/mole 2NH3 -> N2H4 + H2 H = +1876 kJ/mole NOW add, and cancel out terms that appear on both sides. N2 + 2H2 -> N2H4 H = -954 kJ There is another problem. Your ...
Sunday, May 16, 2010 at 12:46am by drwls

Chemistry
You left out reaction arrows N2H4(g) + H2(g) -> 2 NH3(g) H1 = –1876 kJ 3 H2(g) + N2(g) -> 2 NH3(g) H2 = –922 kJ Subtract reaction 2 from 1, and move "-" reactants to the product side: N2H4 -> 2H2 + N2 H3 = -2798 kJ For the reverse (formation) reaction, the heat of ...
Friday, February 22, 2008 at 1:43pm by drwls

chemistry
Dr. Bob, Thank you so much for taking the time to help me. God bless all your work with us. However, I'm still confused b/c these were my choices: 0 mol NH3, 0.038 mol N2, 0.333 mol N2H4, O.150 mol NH3,~0 mol N2, 0.300 mol N2H4, ~0 mol NH3, 0.38 mol N2, 0.188 mol N2H4 and ~ 0 ...
Monday, November 14, 2011 at 11:09am by Lisa

Chemistry
Which pair of samples contains the same number of hydrogen atoms two moles of NH3 and three moles of N2H4 two moles of NH3 and one mole of N2H4 four moles of NH3 and three moles of N2H4 one mole of NH3 and one mole of N2H4
Wednesday, June 29, 2011 at 6:26pm by Brandon

Chemistry
Consider this equilibrium N2(g) + H2(g) <==> NH3(g) +94 kJ The equilibrium law exoression for the balanced chemical equationwould be A. [N2][H2]/[NH3] B. [NH3]/[H2][N2] C. [NH3]2/[H2][N2] D. [NH3]2/[H2]3[N2] E. 2[NH3]2/3[H2]3[N2]
Tuesday, January 3, 2012 at 12:10pm by Q

College Chemistry
In a reaction vessel, the following reaction was carried out using 0.250 mol of NH3(l) and 0.100 mol of N2(g). 4NH3(l) + N2(g) ¡ú 3N2H4(l) Which of the following represents the composition in moles in the vessel when the reaction reaches completion? ¡Ö0 mol NH3, 0.038 mol N2, ...
Monday, March 26, 2012 at 9:23pm by Lisa

Honors Chemistry
"How many grams of NH3 can be produced from the reaction of 28 g of N2 and 25 g of H2? N2 + 3H2 ---> 2NH3 ==> 28 g N2 x (1 mol N2/14.0 g N2) x (2 mol NH3/1 mol N2) x ( 17.0 g NH3/ 1 mol NH3) = 68 g NH3 25 g H2 x ( 1mol H2/2.0 g H2) x (2 mol NH3/3 mol H2) x (17 g NH3/2 ...
Sunday, November 8, 2009 at 8:39pm by Emily

Chemistry
Hydrazine, N2H4, is a colorless liquid used in rocket fuel. What is the enthalpy change for the process in which hydranzine is formed from its elements? N2(g) + H2(g) => N2H4(l) Use the following reactions and enthalpy changes: N2H4 + O2 => 2H2O DeltaH = -622.2KJ H2 + 1/...
Wednesday, September 10, 2008 at 2:35am by Austin

Chemistry
srry I placed the question in the wrong spot. The equation is 2NH3 <=> N2 + 3H2 1 mole of NH3 injected intoa 1L flask 0.3 moles of H2 was found the concentration of N2 at equlibrium is 0.1M how do I find the concentration of NH3 at equlibrium? (N2) = 0.1 (H2) = 0.3 ...
Tuesday, May 8, 2007 at 6:14pm by Bethany

CHEMISTRY
Method for how to get answer? Thank you! For the equilibrium reaction below. To what is Kc equal? N2(g) + 3H2(g) ↔ 2NH3(g) a. [N2] + 3[H2] ------------------ 2[NH3] b. [N2][H2]3 ------------------ [NH3]2 c. 2[NH3] --------------- [N2]+ 3[H2] d. [NH3]2 --------------- [N2...
Sunday, April 6, 2014 at 12:08pm by Chemistry

chemistry
Nitrogen (N2) and hydrogen (H2) react to form ammonia (NH3). Consider a mixture of six nitrogen molecules and six hydrogen molecules in a closed container. Assuming the reaction goes to completion, what will the final product mixture be? number of NH3 molecules number of N2 ...
Friday, January 23, 2009 at 6:35pm by lyne

chemistry
#1. Write the equation and balance it. N2 + 3H2 ==> 2NH3 initial concn. N2 = 6 N2 molecules H2 = 6 H2 molecules NH3 = 0 molecules. If it goes to completion, one of these must be the limiting reagent. How much NH3 would 6 molecules N2 produce? That will be 6 molecules N2 x (...
Friday, January 23, 2009 at 6:35pm by DrBob222

chemistry
First, we write the balanced chemical equation involved. Since the reaction is synthesis of NH3, N2 + 3 H2 -> 2 NH3 Since N2 is excess, H2 is limiting and we therefore use its moles to determine the amount of NH3 produced. In the equation, there are 2 moles of NH3 produced ...
Saturday, September 14, 2013 at 6:59pm by Jai

Chemistry
N2 + 3 H2 -> 2 NH3 Just use their stoichiometric coefficients (number before the chemical's name) from the balanced chemical reaction to make ratios and solve for the unknown. In this case, 2 moles of NH3 are produced for every 3 moles of H2. Thus, 7.5 moles H2 x (2 moles ...
Monday, March 18, 2013 at 7:28am by Jai

Chemistry
Calculate Activation Energy, Ea, for each of the reactions, above? ln(k)= -2.24x10^4(1/T)-1.8. The reactions are Step 1: 2H2 + N2 -> N2H4, Step 2: N2H4 + H2 -> 2NH3, Overall: 3H2 + N2 -> NH3. This is all the information i have. i Don't know how to solve for the ...
Tuesday, June 4, 2013 at 10:23am by Max

Honors Chemistry
You are correct that N2 is the limiting reagent but your numbers are not correct. N2 + 3H2 ==> 2NH3 moles N2 = 28/28 = 1.0 moles NH3 = moles N2 x (2 moles NH3/1 mol N2) = 1.0 x 2 = 2.0 grams NH3 = 2.0 x 17 = 34 g. moles H2 = 25/2 = 12.5 moles. moles NH3 produced = 12.5 x (2...
Sunday, November 8, 2009 at 8:39pm by DrBob222

stoichiometry
first, make sure the equation is balanced,, to determine which is limiting, we calculate the amount of product produced by each given: for 6 mol N2: 6 mol N2 * (2 mol NH3 / 1 mol N2) = 12 mol NH3 for 12 mol H2: 12 mol H2 * (2 mol NH3 / 3 mol H2) = 8 mol NH3 (a) since H2 ...
Wednesday, March 23, 2011 at 11:00pm by Jai

stoichiometry
**i reposted this** first, make sure the equation is balanced,, to determine which is limiting, we calculate the amount of product produced by each given: for 6 mol N2: 6 mol N2 * (2 mol NH3 / 1 mol N2) = 12 mol NH3 for 12 mol H2: 12 mol H2 * (2 mol NH3 / 3 mol H2) = 8 mol NH3...
Wednesday, March 23, 2011 at 10:59pm by Jai

chemistry
You do these with a factor. The right factor will convert anything to anything. Here is how you do this one. But I caution you to use caps when required. h2 and n2 don't mean anything to anyone. Also note that you have asked a limiting reagent question. I know that because you...
Sunday, July 1, 2012 at 1:54pm by DrBob222

Chemistry
N2 + 3H2 => 2NH3 mols H2 = 1.90 mols H2 x (2 mols NH3/3 mol H2) = 1.90 x 2/3 = ? mol NH3. Then g = mols NH3 x molar mass NH3.
Tuesday, March 5, 2013 at 3:54pm by DrBob222

chemistry
Calculate the delta H for the reaction N2H4(l)+ O2(g) --> N2(g) + 2H20(l) Given the following data: 2NH3(g) + 3N20(g) --> 4N2(g) + 3H20(l) N20(g) + 3H2(g) --> N2H4(l) + H20(l) 2NH3(g) + 1/2O2(g) --> N2H4(l) + H20(l) H2(g) + 1/2O2(g) --> H20(l)
Wednesday, October 28, 2009 at 4:35pm by kilo

Chemistry
You use the coefficients just as you did in dimensional analysis. N2 + 3H2 ==> 2NH3. What volume of NH3 will be formed with 1 g hydrogen. 1 mole H2 has a mass of 2 grams and it has a volume of 22.4L. Therefore, 1 g hydrogen will have a volume of 11.2 L (22.4L/mol) x (1 g) x...
Tuesday, March 1, 2011 at 9:03pm by DrBob222

Chemistry
3H2 + N2 ==> 2NH3 Convert 2.90 moles N2 to moles NH3 using the coefficients in the balanced equation. 2.90 x (2 moles NH3/1 mole N2) = 2.90 x 2/1 = ?? Convert moles NH3 to grams by g = moles x molar mass. b.1. Convert 13.75 g NH3 to moles. moles = grams/molar mass = 13.75/...
Monday, October 5, 2009 at 8:03pm by DrBob222

Chem
Correct the equation to N2 + 3H2 ==> 2NH3 Set up an ICE chart. initial: N2 = 1 mole/2 L = 0.5 M H2 = 3 moles/2 L = 1.5 M NH3 = 0 change: NH3 = +2x N2 = -x H2 = -3x equilibrium: N2 = 0.5-x H2 = 1.5-3x NH3 = 2x Substitute into Kc expression and solve. Post your work if you ...
Wednesday, April 14, 2010 at 5:23pm by DrBob222

Chemistry
Determine the number of grams of NH3 produced by the reaction of 3.5g of hydrogen gas with sufficient nitrogen gas. All of these stoichiometry problems are work almost the same. Sometimes an extra step or two are required. Here are the basics. Step 1. Write the balanced ...
Tuesday, January 23, 2007 at 11:04am by Sara

Chemistry
The problem in 1 says that it proceeds to "completion" which means to me that we don't worry about the Kc or Kp. N2 + 3H2 ==> 2NH3. Since both reactants are given we know it is a limiting reagent problem. This is a gas problem; therefore, we are allowed to use L directly as...
Wednesday, August 10, 2011 at 3:33pm by DrBob222

Chemistry, Need Help!!
a)How many grams of H2 are needed to produce 14.27 g of NH3? b)How many molecules (not moles) of NH3 are produced from 1.57×10−4 g of H2 ?
Monday, October 3, 2011 at 11:03pm by Hannah

Chemistry!
Write the equation and balance it. Convert 90 g H2 to mols. mols = g/molar mass. Convert mols H2 to mols NH3 by using the coefficients in the balanced equation. Convert mols NH3 to grams NH3. g NH3 = mols NH3 x molar mass NH3. That is the theoretical yield.
Tuesday, November 13, 2007 at 3:30pm by DrBob222

chemistry
I'm a little confused by the question; I may write more than is necessary. This question is about Le Chatelier's principle and which direction the reaction will shift. Shift to the right means more NH3 is produced at the expense of N2 and H2. Shift to the left means NH3 is ...
Monday, February 18, 2013 at 4:44pm by DrBob222

Chemistry
N2 + 2H2 ==> N2H4 dGrxn = (n*dGf products) - (n*dGf reactants). According to my tables N2H4 is -16 kJ/mol, N2 and H2 are zero. You need to use the numbers in your tables.
Thursday, March 7, 2013 at 11:22pm by DrBob222

chem- i reallyneed help
Consider the following equations. N2H4(l) + O2(g) N2(g) + 2 H2O(l) ÄH = -622.2 kJ H2(g) + 1/2 O2(g) H2O(l) ÄH = -258.5 kJ H2(g) + O2(g) H2O2(l) ÄH = -187.8 kJ Use this information to calculate the enthalpy change for the reaction shown below. N2H4(l) + 2 H2O2(l) N2(g) + 4 H2O(...
Tuesday, November 6, 2012 at 10:30am by hannah

Chemistry
initial pressure NH3 = 0.7317 atm. initial pressure H2 = 0 and N2 = 0 At equilibrium N2 = x H2 = 3x NH3 = 0.7317 - 2x Kp = 67,100 = pN2*pH2^3/pNH3^2 Substitute into Kp and solve for x, then convert to NH3, H2, N2. Finally, pNH3 remaining (from above) to percent of original ...
Sunday, October 3, 2010 at 9:14pm by DrBob222

Honors Chemistry
No, no, and no. 28 g N2 = ??moles = 28/14 = 2 moles N2. 2 moles N2 x (2 moles NH3/1 mole N2) = 2*2/1 = 4 moles NH3. grams = moles NH3 x molar mass NH3 = about close to 70 g. There is NO NH2 produced. It isn't even in the equation. How much H2 remains un-reacted? You have 25 g ...
Friday, January 28, 2011 at 8:45pm by DrBob222

Chemistry
So I have had several homework problem like this and managed to solve them easily, but these two problems I cannot get the correct answer for. Using a Table of thermodynamic data, calculate the change in Gibbs free energy for each of the following reactions. In each case ...
Thursday, February 17, 2011 at 12:06am by Don

please check my answer
Consider the following equations. N2H4(l) + O2(g) N2(g) + 2 H2O(l) ÄH = -622.2 kJ H2(g) + 1/2 O2(g) H2O(l) ÄH = -258.5 kJ H2(g) + O2(g) H2O2(l) ÄH = -187.8 kJ Use this information to calculate the enthalpy change for the reaction shown below. N2H4(l) + 2 H2O2(l) N2(g) + 4 H2O(...
Tuesday, November 6, 2012 at 9:34pm by hannah

chemistry-Thermochemistry (grade 12)
calculate enthalpy of H for the reaction N2H4(l) + 2H2O(l) -> N2(g) + 4H2)(l) Given the reactions N2H4(l) + O2(g) -> N2(g) + 2H2O(l) Enthalpy of H = -6.22.2 kJ H2(g) + (1/2)O2(g) -> H2O(l) enthalpy of H = -285.8 kJ/mol H2(g) + O2(g) -> H2O2(l) enthalpy of H = -187....
Thursday, February 16, 2012 at 1:14am by Rose Bud

chemistry-Thermochemistry (grade 12)
calculate enthalpy of H for the reaction N2H4(l) + 2H2O(l) -> N2(g) + 4H2)(l) Given the reactions N2H4(l) + O2(g) -> N2(g) + 2H2O(l) Enthalpy of H = -6.22.2 kJ H2(g) + (1/2)O2(g) -> H2O(l) enthalpy of H = -285.8 kJ/mol H2(g) + O2(g) -> H2O2(l) enthalpy of H = -187....
Thursday, February 16, 2012 at 1:36pm by Rose Bud

chemistry
I assume a jugful means an excess of N2. mol H2 = grams/molar mass Using the coefficients in the balanced equation, convert mols H2 to mols NH3. Now convert mols NH3 to grams. g = mols NH3 x molar mass NH3.
Sunday, February 2, 2014 at 12:08pm by DrBob222

Chemistry
N2 + 3H2 ==> 2NH3 Use dimensional analysis and the coefficients. for N2. 6 mol N2 x (2 moles NH3/1 mole N2) = 12 mol NH3 produced. For H2. 12 mol H2 x (2 moles NH3/3 moles H2) = 8 moles NH3 produced. Therefore, hydrogen is the limiting reagent and N2 is in excess because in...
Sunday, February 27, 2011 at 4:48pm by DrBob222

chemistry
determine the number of grams of NH3 produced by the reaction of 3.5g of hydrogen gas. 1. Write the equation and balance it. N2 + 3H2 ==> 2NH3 2. Convert 3.5 g H2 gas to mols. mols = grams/molar mass. 3. Convert mols H2 gas to mols NH3 using the equation from step 1. 4. ...
Saturday, August 11, 2007 at 12:18am by danni

science
The balanced reaction for that process is N2 + 3 H2 = 2 NH3 12.5 g of H2 is 6.25 moles. You will form 2/3 as many moles of NH3. Each mole is 6.02*10^23 molecules. You finish up.
Monday, November 9, 2009 at 9:45am by drwls

chemistry
5. The industrial production of hydroiodic acid takes place by treatment of iodine with hydrazine (N2H4): 2I2 + N2H4 → 4HI + N2 a.How many grams of I2 are needed to react with 36.7 g of N2H4
Monday, November 28, 2011 at 4:22pm by davian

chem
A limiting reagent problem with an extra twist at the end. Write and balance the equation N2 + 3H2 ==> 2NH3 Using the coefficients in the balanced equation, convert mols N2 and mols H2 to mols NH3. N2 first: 1.41 mol N2 x (2 mol NH3/1 mol N2) = 2.82 mols NH3 produced (if we...
Saturday, September 21, 2013 at 5:46pm by DrBob222

Chemisty
This is a limiting reagent problem. I work these the long way. First we determine which is the limiting reagent. How much NH3 can be produced with 0.150 mol N2 and all the H2 needed? 0.150 mol N2 x (2 mol NH3/1 mol N2) = ? How much NH3 can be produced with 0477 mol H2 and all ...
Friday, April 20, 2012 at 6:16am by DrBob222

chemistry(oxidation)
ok,I really dont understand the whole oxidation numbers thing my question is state the oxidation number for nitrogen in the following NH3= -3 NO2-1= -1 N2= 0 NO2Cl= 5 N2H4 -2 am I correct? NH3= -3 NO2-1= -1 No, oxygen is -2, so N is +3 N2= 0 NO2Cl= 5 N2H4 -2 http://www....
Thursday, May 17, 2007 at 10:13am by mike

chemistry
At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction. 2 NH3(g) N2(g) + 3 H2(g) At equilibrium, 2.0 mol NH3 remains. What is the value of K for this reaction?
Sunday, January 8, 2012 at 3:37pm by Franz

AP Chem
At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction. 2 NH3(g) N2(g) + 3 H2(g) At equilibrium, 2.0 mol NH3 remains. What is the value of K for this reaction?
Sunday, January 8, 2012 at 6:25pm by Bill

chem
637 kJ/4 mols H2. So 637/4 = ?? for 1 mol H2. For #2, 637 kJ/mol NH3 x [10 g NH3 x (1 mol NH3/17 g NH3)] = xx
Saturday, March 8, 2008 at 8:39pm by DrBob222

chemistry first year
This is a limiting reagent (LR) problem. You can tell because amounts are give for both reactants. V is voltage, value, volume. Volume maybe. ........N2 + 3H2 ==> 2NH3 begin volume = 20 + 100 = 120 L. What volume NH3 will be produced if we use all of the N2 and and excess ...
Saturday, April 19, 2014 at 6:14am by DrBob222

chemistry
What about the product? Is that N2H4 or NH3.
Monday, November 5, 2012 at 11:24am by DrBob222

chemistry
In the reaction of N2 and H2 to produce NH3, how many moles of H2 will produce 37.9 grams NH3 if sufficient N2 is present? Do not enter units with your answer.
Friday, October 21, 2011 at 12:48am by Charles

chemistry
Start with the balanced reaction, you did not specify it. There are a number of reactions with H2 to make ammonia. Lets assume the Haber process. 3H2 + N2 >> 2NH3 so it appears you get 2/3 the moles H2 when making NH3 2/3 * 1.9= you do it. Now convert that to grams of NH3
Saturday, October 23, 2010 at 11:44am by bobpursley

Chemistry
Could someone please check my answers/work for the following questions. EQUATION: 3H2+N2 -->2NH3 1) How many grams of NH3 can be produced from 3.98 mol of N2? 3.98 mol N2(2 mol NH3/2 mol H2)=2.65 mol Convert to grams: 2.65(17.0)=45.14g 2.)How many grams of H2 are needed to ...
Tuesday, September 28, 2010 at 12:10pm by Sarene

chemistry
Write the equation and balance it. Convert 4.56E-4 g H2 to moles. mols = grams/molar mass Use the coefficients in the balanced equation to convert mols H2 to mols NH3. Convert mols NH3 to molecules. Remember 1 mol NH3 contains 6.02E23 molecules.
Saturday, March 3, 2012 at 3:12am by DrBob222

Chemistry
well one mole of NH3 has 3 moles of H 1 mole of H2 has 2 moles of H each mole of NH3 requires 3/2 moles of H2 mol wt of NH3 = 14 + 3*1 = 17 so you need 3/2 * (13.27 / 17) = 1.17
Wednesday, September 19, 2012 at 1:35pm by Steve

math
H2 : NH3 = 2*1.008 : 14.007+3*1.008 = 2.016 : 17.031 So if H2 weighs 15 g, let x=NH3 produced, cross multiply to get mass of NH3 = 15*17.031/2.016=?
Monday, June 27, 2011 at 12:55pm by MathMate

Chemistry
There is a long way and a shortcut way. First the long way which will work on all problems. This is a simple (as opposed to limiting reagent stoichiometry problem) stoichiometry problem. (USUALLY we know it is a limiting reagent problem when amounts for BOTH reactants are ...
Saturday, November 3, 2012 at 2:37pm by DrBob222

Chemistry
3 moles of H2 produce 2 moles of NH3 3.65*10^-4 g of H2 is 3.621*10^-4 moles H2 That will produce 2.414*10^-4 moles of NH3 2.414*10^-4 * 6.023*10^23 = 1.454*20^20 molecules NH3
Wednesday, February 19, 2014 at 1:01pm by Steve

Chemistry
You are given the following decomposition reaction of ammonia NH3 ---> H2 + N3 unbalanced When you decompose 12.0g of NH3you produce 1.87 g of H2 Calculate the percent yield of H2
Thursday, April 18, 2013 at 2:07am by Johny

Chemistry
A student ran the following reaction in the laboratory at 651 K: 2 NH3(g) N2(g) + 3 H2(g) When she introduced 8.75E-2 moles of NH3(g) into a 1.00 Liter container, she found the equilibrium concentration of H2(g) to be 0.119 M. Calculate the equilibrium constant, Kc, she ...
Tuesday, November 19, 2013 at 10:50pm by Jessica

chemistry
I assume we are talking about: 3 H2 + N2 --> 2 NH3 H2 is 2*1 = 2 g/mol N2 is 2*14 = 28 g/mol forget g vs kg because we are only interested in ratios here 3 mol H2 is 6 g so we need 6 g of H2 for every 28 g of N2 31.5 g of N2 needs 31.5 * 6/28 = 6.75 g H2 needed BUT we only ...
Thursday, October 18, 2012 at 6:05am by Damon

Science!
How many grams of NH3 gas can be made if 42.0 g of N2 gas and 12.0 g of H2 gas react with each other? N2 + 3 H2 ↔ 2 NH3
Monday, February 11, 2013 at 12:42am by Jacob

Chemistry
N2(g) + H2(g) -> NH3(g) ; how many moles of NH3 is is produced when 0.67 moles of N2 reacts with H2?
Sunday, March 4, 2012 at 7:49pm by Suzanne

chemistry
If you increase the N2 or H2 concentrations that is going to force the production of NH3 If you increase the temperature, to maintain equilibrium heat has to be given off. Which means more NH3 If you add pressure you need fewer molecules which again causes the production of ...
Monday, February 18, 2013 at 5:04pm by JJ

chemistry
A mixture of hydrogen peroxide, H2O2, and hydrazine, N2H4, can be used as a rocket propellant. The reaction is: 7 H2O2(g) + N2H4(l) ® 2 HNO3(aq) + 8 H2O(l) a) How many moles of H2O2 react with 0.477 mol N2H4? [1] ___________ b) How many grams of HNO3 can be produced in a ...
Saturday, June 20, 2009 at 12:41am by dan

Chemistry..
A chemical engineer calculated that 15.0 mol H2 was needed to react with excess N2 to prepare 10.0 mol NH3. But the actual yield is 60.0% . Write a balanced chemical equation for the reaction. Is the amount of H2 needed to make 10.0 mol NH3 more than, the same as, or less than...
Wednesday, January 2, 2013 at 8:46pm by Anonymous

chemistry
I's 0.50, not 0.05. The short answer to your question is that I didn't. Don't confuse the mols given in the problem with the value of x in the solution. 1.5 mols is given in the problem as mols H2 at equilibrium; therefore, if it had asked for (H2) we would have written 1.5mol...
Saturday, May 5, 2012 at 4:12pm by DrBob222

10th grade chemistry
When everything is in the gaseous state, one may use a short cut; i.e., it isn't necessary to convert to mols of what you have, then to mols of what you want, and back to moles of what you want. Go directly from L of what you have to L of what you want. 25.0L H2 gas x (2 moles...
Tuesday, May 29, 2012 at 6:27pm by DrBob222

Chemistry Honors
I assume that is a 2.0L container but you omitted the unit. M NH3 = mols/L = 12.0/2.0 = 6.0M ........2NH3 ==> N2 + 3H2 I......6.0.......0.....0 C......-2x.......x.....3x E......6.0-2x....x.....3x Kc = (N2)(H2)^3/(NH3)^2 The problems tells you H2 at equil is 6.0mols; ...
Saturday, July 20, 2013 at 5:20pm by DrBob222

Chemistry
L N2 = 1.19E4L NH3 x (1/2) = ? L H2 = 1.19E4L NH3 x (3/2) = ? Total V at same T and P is the sum L N2 + L H2.
Saturday, October 26, 2013 at 11:02pm by DrBob222

Chemistry
Write and balance the equation. 3H2 + N2 ==> 2NH3 Using the coefficients in the balanced equation, convert mols H2 to mols NH3. Then convert mols NH3 to g. grams = mols NH3 x molar mass NH3.
Wednesday, October 31, 2012 at 9:19pm by DrBob222

Chemistry
In the industrial synthesis of ammonia, the equilibrium constant expression may be written as: Keq= [NH3]^2/[N2][H2]^3 Calculate the value of this equilibrium constant, if the equilibrium concentration of nitrogen in the reaction mixture at 600°C if [N2] = 4.53 and [H2] = 2.49...
Friday, November 18, 2011 at 12:24am by Chemistry Chick

Chem 111
if the equilibrium concentration of nitrogen at 18degrees C is 0.256M and equil. concen. of hydrogen is 0.0184M what is the equil. concen. of ammonia? N2 + 3 H2 <=> 2 NH3; so K= [NH3]^2 / [N2][H2]^3 Is this correct? Then ICE table? Giving me.... K= (2x)^2 / (0.256-x)(0....
Thursday, September 8, 2011 at 12:25pm by Anonymous

Chemistry
Balance the equation. Convert 8.50 g N2 to mols. Mols = grams/molar mass. Convert 1.00 g H2 to mols. Using the coefficients in the balanced equation, convert mols H2 and mols N2 to mols NH3. The correct answer for mols NH3 will be the smaller of the two answera you obtain. ...
Thursday, October 4, 2007 at 6:23pm by DrBob222

Chem
For each unbalanced equation below, calculate how many grams of each product would be produced by complete reaction of 12.5g of the reactant indicated. Indicate clearly the mole ratio used for the conversion. A. TiBr4(g)+H2(g)=Ti(s)+HBr(g), Reactant is H2 B. SiH4(g)+NH3(g)=...
Wednesday, June 25, 2008 at 2:59pm by OliviaK

college chem
The cation M2+ reacts with NH3 to form a series of complex ions as follows: M2+ + NH3 M(NH3)2+ K1 = 102 M(NH3)2+ + NH3 M(NH3)2 2+ K2 = 103 M(NH3)2 2+ + NH3 M(NH3)3 2+ K3 = 102 A 1.0 × 10–3 mol sample of M(NO3)2 is added to 1.0 L of 15.0 M NH3 (Kb = 1.8 × 10–5). What is the ...
Sunday, April 24, 2011 at 7:19pm by richard

CHEMISTRY
Do you mean at equilibrium (NH3) = 1.23E-4M? and (H2) at equilibrium = 2.75E-3M? that is, is that M? ........N2 + 3H2 ==> 2NH3 E.......x....2.75E-3....1.23E-4M Keq = (NH3)^2/(x)(H2)^3 Substitute and solve for x
Thursday, April 10, 2014 at 11:20pm by DrBob222

Chemistry
All are incorrect for the same reason. Everything is ok except the conversion factors. 1) How many grams of NH3 can be produced from 3.98 mol of N2? 3.98 mol N2(2 mol NH3/2 mol H2)=2.65 mol Convert to grams: 2.65(17.0)=45.14g 3.98 mol N2 x (2 mol NH3/1 mol N2) = 7.96 mol NH3. ...
Tuesday, September 28, 2010 at 12:10pm by DrBob222

please help w/ chem hw
Kc = [NH3]^2 / [N2] [H2]^3 so [N2] = [NH3]^2 / [H2]^3 Kc
Sunday, July 28, 2013 at 10:33pm by Graham

chemistry
Hydrogen cyanide is a highly poisonous, volatile liquid. It can be prepared by the following reaction. CH4(g) + NH3(g) → HCN(g) + 3 H2(g) What is the heat of reaction at constant pressure? Use the following thermochemical equations. N2(g) + 3 H2(g) → 2 NH3(g) Δ...
Friday, October 29, 2010 at 12:48am by caroline

Chemistry
Hydrogen cyanide is a highly poisonous, volatile liquid. It can be prepared by the following reaction. CH4(g) + NH3(g) → HCN(g) + 3 H2(g) What is the heat of reaction at constant pressure? Use the following thermochemical equations. N2(g) + 3 H2(g) → 2 NH3(g) Δ...
Sunday, October 31, 2010 at 4:05pm by Emily

Chemistry
What volume of hydrogen is necessary to react with five liters of nitrogen to produce ammonia? (assume constant temprature and pressure) Balanced equation- N2 + 3H2= 2NH3 After finding this answer, what volum of ammonia is produced in this reaction? 1. Write the equation. You ...
Sunday, January 28, 2007 at 12:58pm by Kylen

chemistry
You didn't balance th equation and ammonia is NH3. N2H4 is hydrazine. Did you intend to write concns (and not partial pressures) for the reactants and products.
Monday, November 5, 2012 at 11:24am by DrBob222

Chemistry
Given the balanced equation N2(g)+3H2(g)==>2NH3 Calculate: a)the volume H2 that reacts with 12L of N2 b)the volume of NH3 produced Iron 4L ofN2 c)the volume of N2 and H2 to produced 60L of NH3 Assume that all volume measurements are made under identical conditions
Monday, April 23, 2012 at 12:59pm by faisal

chem, find concentrations
A 2 liter container holds 2.2 mol of NH3 gas, which starts to decompose according to the following reaction. At equilibrium there are .6 mol of H2. Determine the concentrations of NH3, H2, and N2 at equilibrium. 2NH3-> 3H2+N2 ^ all gas
Monday, May 14, 2012 at 11:28pm by bkue

Chemistry
N2 + 3H2 ==> 2NH3 When using equations will all gases, one may use L as if they were moles. 1.5L N2 will produce 1.5 x (2/1) = 3.0L NH3 and 4.5L H2 will produce 4.5 x (2/3) = 3.0 L NH3. So the answers is that 3.0 L NH3 will be produced at STP.
Tuesday, November 1, 2011 at 1:17pm by DrBob222

chemistry
N2(g)+3H2(g)<=>2NH3(g) +▲H if uou are preparing NH3, hiw does the following increase the yield of NH3? (a) ▲[N2] ▲[H2] (b) ▲T of the chamber storing NH3 (c) ▲P of the system (d) ▲[NH3(g)] * <=> equilibrium * ▲ change
Monday, February 18, 2013 at 4:39pm by gaby

chemistry
N2(g)+3H2(g)<=>2NH3(g) +▲H if uou are preparing NH3, hiw does the following increase the yield of NH3? (a) ▲[N2] ▲[H2] (b) ▲T of the chamber storing NH3 (c) ▲P of the system (d) ▲[NH3(g)] * <=> equilibrium * ▲ change
Monday, February 18, 2013 at 4:44pm by gaby

chemistry
N2(g)+3H2(g)<=>2NH3(g) +▲H if uou are preparing NH3, hiw does the following increase the yield of NH3? (a) ▲[N2] ▲[H2] (b) ▲T of the chamber storing NH3 (c) ▲P of the system (d) ▲[NH3(g)] * <=> equilibrium * ▲ change
Monday, February 18, 2013 at 5:04pm by gaby

Chemistry
Which of the gases most closely resembles an ideal gas at standard temperature and pressure? 1) CO2 2) NH3 3) HI 4) H2 5) H2O why would it be H2
Sunday, August 29, 2010 at 10:37pm by Amy~

Chemistry
initial moles: NH3 = 10 N2 = 0 H2 = 0 equilibrium: N2 = 1.2 moles from problem. Therefore, H2 must be 3 x 1.2 = 3.6 NH3 must be 10-(2*1.2) = ??
Tuesday, June 8, 2010 at 5:20pm by DrBob222

Chemistry
Write the Kc expression from the equation.. N2 + H2 ==> 2NH3 Then 0.0186 M NH3 initially. At equilibrium H2 must be 3x, N2 must be 2x, and NH3 must be 0.0186-2x. Plug those values in and solve for Kc.
Saturday, November 3, 2007 at 6:15pm by DrBob222

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