Hydrogen cyanide, HCN, can be made by a two-step process. First, ammonia is reacted with O2 to give nitric oxide, NO.
4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Then nitric oxide is reacted with methane, CH4.
2 NO(g) + 2 CH4(g) → 2 HCN(g) + 2 H2O(g) + H2(g)
When 28.6 g of ammonia and 25.1 g of methane are used, how many grams of hydrogen cyanide can be produced?
CHEMISTRY - DrBob222, Sunday, September 27, 2009 at 7:21pm
This is a limiting reagent type problem. All of them can be worked in the following way.
1. Write and balance the equation. You have that.
2a. Convert 28.6 g NH3 to moles. Moles = grams/molar mass.
2b. Convert 25.1 g CH4 to moles.
3a. Using the coefficients in the balanced equation, convert moles NH3 to moles HCN.
3b. Using the coefficients in the balanced equation, convert moles CH4 to moles HCN.
3c. Obviously, one of the answers (either 3a or 3b) must be wrong. The smaller one is ALWAYS the correct value to use and the reagent that produced that value is the limiting reagent.
4. Using the smaller value from 3c, convert moles of HCN to grams. grams = moles x molar mass.
Post your work if you get stuck.
CHEMISTRY - Jason, Sunday, September 27, 2009 at 9:20pm
I don't understand 3a and 3b. I don't understand how you do that
CHEMISTRY - DrBob222, Sunday, September 27, 2009 at 11:47pm
OK. For 2a and 2b, we have the following but I have just used round numbers for the molar masses. You may want to redo them with more exact values.
28.6/17 = 1.68 moles NH3.
25.1/16 = 1.57 moles CH4.
I have led you astray a little. The NH3 must be converted to NO first. So we want to convert 1.68 moles NH3 to moles NO. Same directions.
moles NO = 1.68 moles NH3 x (4 moles NO/4 moles NH3) = 1.68 x (4/4) = 1.68 x 1/1 = 1.68 moles NO.[Note that the unit moles NH3 cancel and converts moles NH3 to moles NO. Now we convert moles NO to moles HCN.
moles HCN = 1.68 moles NO x (2 moles HCN/2 moles NO) = 1.68 x (2/2) = 1.68 x 1/1 = 1.68 moles HCN.
3b. Convert moles CH4 to moles HCN.
moles HCN = 1.57 moles CH4 x (2 moles HCN/2 moles CH4) = 1.57 x (2/2) = 1.57 x (1/1) = 1.57 moles HCN.
3c. The smaller number rules so CH4 is the limiting reagent and 1.57 moles HCN will be produced.
4. Now we convert 1.57 moles HCN to grams.
1.57 moles HCN x molar mass HCN = grams HCN.
Sorry about the mixup. I hope this is clear. If not, I think it would be better to start a new post at the top of the page, recopy the problem and post additional questions.
CHEMISTRY - sally, Thursday, September 23, 2010 at 11:45pm
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CHEMISTRY - billy, Thursday, September 23, 2010 at 11:46pm
sally is a big poo headed jerk face. she is sitting beside me drinking cranberry juice and just being a jerk. it's an everyday thing for her. you have to understand one thing... she's a playstation girl.... enough said.
CHEMISTRY - sally, Thursday, September 23, 2010 at 11:49pm
for one thing it is cran-grape, mister. secondly, playstation kills xbox. the end :) oh and i'm the nicest girl ever. i mean, come on, my name is sally :D
CHEMISTRY - billy, Thursday, September 23, 2010 at 11:54pm
cran-grape, shmamgrape... it all taste like poooooooooop. and yea ok miss i play a ps2. GET IN THE 21st CENTURY!!! and actually sally means baby eating booger flicker sooo ummm.... no. and o yea, your mac sucks..
CHEMISTRY - sally, Friday, September 24, 2010 at 12:03am
pc's suck :)
CHEMISTRY - jennifer, Tuesday, February 8, 2011 at 5:13pm
When you react hydrogen gas with nitrogen gas to form ammonia, How many grams of all species ( reactants and products) are left when you start with 140 grams of nitrogen and 40 grams of hydrogen?
CHEMISTRY - jennifer, Tuesday, February 8, 2011 at 5:17pm
Calculate the mass of water produced by 64 grams of CH4
CH4(G) - 2O2 (G) -----CO2(G)-2H2O(l)