Consider the following reaction: 4NH3 + 7O2 (arrow) 4NO2 + 6H2O.
A) How many moles of NH3 react with 5.64 mol of O2?
B)How many moles of NO2 are obtained from 3.27 mol of O2?
C) How many moles H2O will be produced from 8.95 g of NH3?
D)How many grams of NH3 will be needed to produce 0.0160 g NO2?
Please show work and explain how you got them!! I am lost and I have 10 problems like this!!
Two rules to follow in chemistry problems.
1. Work in mols. You get mols two ways. The first is when you have grams; that's mols = grams/molar mass. The second is when you have a solution; then mols = M x L.
2. When you have mols you can convert mols of anything in the equation to mols of anything else in the equation by using the coefficients in the balanced equation.
a. 5.84 mols O2 x (4 mols NH3/7 mols O2) = 5.84 x 4/7 = ?
b. same procedure as in a. Note that the factor you use (in the above case 4/7) cancels the mols O2 unit and converts to mols NH3 unit. In these conversions the unit you don't want to keep always goes in the denominator of the fraction and the unit you want to convert to goes in the numerator of the fraction.
c. When you have grams, first convert to mols.
mols NH3 = grams/molar mass = 8.95/17 \ estimated 0.53. Then convert to mols of what you want.
0.53 mols NH3 x (6 mols H2O/4 mol NH3) = 0.53 x (3/2) = estimated 0.79 mols H2O. The problem doesn't ask you to convert to grams (but the next one does). To find grams just g = mols x molar mass = about 0.79 x 18 = ?
A) To determine the number of moles of NH3 that react with 5.64 mol of O2, we need to identify the stoichiometric ratio between NH3 and O2 in the balanced equation. According to the balanced equation, 4 moles of NH3 react with 7 moles of O2.
Therefore, we can set up a proportion:
4 moles NH3 / 7 moles O2 = x moles NH3 / 5.64 moles O2
Cross-multiplying and solving for x:
4 moles NH3 * 5.64 moles O2 = 7 moles O2 * x moles NH3
22.56 = 7x
Dividing both sides by 7:
x = 22.56 / 7
x ≈ 3.223 moles NH3
So, approximately 3.223 moles of NH3 will react with 5.64 mol of O2.
B) Similar to part A, we can set up a proportion to determine the number of moles of NO2 obtained from 3.27 mol of O2. According to the balanced equation, 7 moles of O2 react and form 4 moles of NO2.
7 moles O2 / 4 moles NO2 = 3.27 moles O2 / x moles NO2
Cross-multiplying and solving for x:
7 moles O2 * x moles NO2 = 4 moles NO2 * 3.27 moles O2
7x = 13.08
Dividing both sides by 7:
x = 13.08 / 7
x ≈ 1.868 moles NO2
Approximately 1.868 moles of NO2 are obtained from 3.27 mol of O2.
C) To determine the number of moles of H2O produced from 8.95 g of NH3, we need to convert the mass of NH3 to moles using its molar mass. The molar mass of NH3 is:
1 mole of N + 3 moles of H = 14.01 g + (3 × 1.01 g) = 17.03 g
Using the molar mass, we can set up a proportion:
17.03 g NH3 / 1 mole NH3 = 8.95 g NH3 / x moles NH3
Cross-multiplying and solving for x:
17.03 g NH3 * x moles NH3 = 8.95 g NH3 * 1 mole NH3
17.03x = 8.95
Dividing both sides by 17.03:
x = 8.95 / 17.03
x ≈ 0.526 moles H2O
Approximately 0.526 moles of H2O will be produced from 8.95 g of NH3.
D) To determine the grams of NH3 needed to produce 0.0160 g of NO2, we can use the molar mass of NH3 (17.03 g/mole) to convert to moles, and then use the stoichiometric ratio from the balanced equation.
First, convert 0.0160 g of NO2 to moles using its molar mass (46.01 g/mole):
0.0160 g NO2 / 46.01 g/mol = 0.000347 moles NO2
Next, set up a proportion using the stoichiometric ratio:
4 moles NH3 / 4 moles NO2 = x moles NH3 / 0.000347 moles NO2
Cross-multiplying and solving for x:
4 moles NH3 * 0.000347 moles NO2 = 4 moles NO2 * x moles NH3
0.001388 = 4x
Dividing both sides by 4:
x = 0.001388 / 4
x ≈ 0.000347 moles NH3
Finally, convert moles of NH3 to grams using its molar mass:
0.000347 moles NH3 * 17.03 g/mol = 0.00592 g NH3
Approximately 0.00592 grams of NH3 will be needed to produce 0.0160 g of NO2.
A) To determine how many moles of NH3 react with 5.64 mol of O2, we need to use the stoichiometric ratio between NH3 and O2 in the balanced equation.
From the balanced equation, we can see that the ratio between NH3 and O2 is 4:7. This means that for every 4 moles of NH3, we need 7 moles of O2.
Using this ratio, we can set up a proportion:
(4 moles NH3) / (7 moles O2) = (x moles NH3) / (5.64 moles O2)
Cross-multiplying and solving for x gives:
(4 moles NH3) * (5.64 moles O2) = (7 moles O2) * (x moles NH3)
x = (4 moles NH3 * 5.64 moles O2) / 7 moles O2
x = 3.22 moles NH3
Therefore, 3.22 moles of NH3 will react with 5.64 mol of O2.
B) To determine how many moles of NO2 are obtained from 3.27 mol of O2, we again use the stoichiometric ratio between O2 and NO2 from the balanced equation.
From the balanced equation, we can see that the ratio between O2 and NO2 is 7:4.
Using this ratio, we can set up a proportion:
(7 moles O2) / (4 moles NO2) = (3.27 moles O2) / (x moles NO2)
Cross-multiplying and solving for x gives:
(7 moles O2) * (x moles NO2) = (4 moles NO2) * (3.27 moles O2)
x = (7 moles O2 * 3.27 moles O2) / 4 moles NO2
x = 5.71 moles NO2
Therefore, 5.71 moles of NO2 are obtained from 3.27 mol of O2.
C) To determine how many moles of H2O will be produced from 8.95 g of NH3, we need to convert grams of NH3 to moles using the molar mass of NH3.
The molar mass of NH3 is calculated as follows:
(Molar mass of N) + 3 * (Molar mass of H)
= (14.01 g/mol) + 3 * (1.01 g/mol)
= 17.04 g/mol
Now, we can set up a conversion:
(8.95 g NH3) * (1 mol NH3 / 17.04 g NH3) * (6 mol H2O / 4 mol NH3)
Simplifying:
(8.95 / 17.04) * (6 / 4) = 2.37 moles H2O
Therefore, 2.37 moles of H2O will be produced from 8.95 g of NH3.
D) To determine how many grams of NH3 will be needed to produce 0.0160 g NO2, we need to use the stoichiometric ratio between NH3 and NO2 from the balanced equation.
From the balanced equation, we can see that the ratio between NH3 and NO2 is 4:4, meaning the ratio is 1:1.
Given that the mass of NO2 is 0.0160 g, the mass of NH3 needed can be found using the molar mass of NH3.
The molar mass of NH3 is 17.04 g/mol. So, we can set up the following conversion:
(0.0160 g NO2) * (1 mol NO2 / 46.0 g NO2) * (1 mol NH3 / 1 mol NO2) * (17.04 g NH3 / 1 mol NH3)
Simplifying:
(0.0160 / 46.0) * (17.04 / 1) = 0.00591 g NH3
Therefore, 0.00591 g of NH3 will be needed to produce 0.0160 g of NO2.
A) To determine the moles of NH3 that react with 5.64 mol of O2, we need to use the stoichiometry of the reaction.
According to the balanced equation, 4 moles of NH3 react with 7 moles of O2.
Step 1: Use the molar ratio from the balanced equation to convert moles of O2 to moles of NH3.
Moles of NH3 = (5.64 mol O2) x (4 mol NH3 / 7 mol O2)
= 3.22 mol NH3
Therefore, 3.22 moles of NH3 react with 5.64 mol of O2.
B) To find the moles of NO2 obtained from 3.27 mol of O2, we will again use the stoichiometry of the reaction.
According to the balanced equation, 4 moles of NH3 react with 4 moles of NO2 and 7 moles of O2.
Step 1: Use the molar ratio from the balanced equation to convert moles of O2 to moles of NO2.
Moles of NO2 = (3.27 mol O2) x (4 mol NO2 / 7 mol O2)
= 1.88 mol NO2
Therefore, 1.88 moles of NO2 are obtained from 3.27 moles of O2.
C) To determine the moles of H2O produced from 8.95 g of NH3, we first need to convert grams of NH3 to moles of NH3, and then use the stoichiometry of the reaction to convert moles of NH3 to moles of H2O.
Step 1: Convert grams of NH3 to moles of NH3 using the molar mass of NH3.
Moles of NH3 = (8.95 g NH3) / (17.03 g/mol NH3)
= 0.526 mol NH3
Step 2: Use the molar ratio from the balanced equation to convert moles of NH3 to moles of H2O.
Moles of H2O = (0.526 mol NH3) x (6 mol H2O / 4 mol NH3)
= 0.789 mol H2O
Therefore, 0.789 moles of H2O will be produced from 8.95 g of NH3.
D) To calculate the grams of NH3 needed to produce 0.0160 g of NO2, we will use the stoichiometry of the reaction.
Step 1: Convert grams of NO2 to moles of NO2 using the molar mass of NO2.
Moles of NO2 = (0.0160 g NO2) / (46.01 g/mol NO2)
= 0.000347 mol NO2
Step 2: Use the molar ratio from the balanced equation to convert moles of NO2 to moles of NH3.
Moles of NH3 = (0.000347 mol NO2) x (4 mol NH3 / 4 mol NO2)
= 0.000347 mol NH3
Step 3: Convert moles of NH3 to grams of NH3 using the molar mass of NH3.
Grams of NH3 = (0.000347 mol NH3) x (17.03 g/mol NH3)
= 0.00590 g NH3
Therefore, 0.00590 grams of NH3 will be needed to produce 0.0160 g of NO2.