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4NH 3 + 5O2 ® 4NO + 6H 2 0
1. How many moles of each reactant are required to produce 4.250 moles of
NO?
4NH 3
4.250 moles of NO
= 4.250 moles NH 3
4NO
4.250 moles of NO
5O2
= 5.313 moles O2
4NO
2. Show two different ways to solve for the number of moles of H2O produced
when 4.250 moles of NO are produced (use numbers from problem #1).
4.250 moles of NO
5.313 moles O2
6H 2O
= 6.375 moles H 2O
4NO
6H 2O
= 6.375 moles H 2O
5O2
4.250 moles of NH3
6H 2O
= 6.375 moles H 2O
4NH 3
3. If 6.214 moles of H2O are produced from 4.250 moles of NO, calculate the
percent yield for the reaction.
4.250 moles of NO should produce 6.375 moles of H2O
6.214
(100) = 97.47 % yield
6.375
3Fe + 4H 2O ® Fe3O4 + 4H 2
4. If 100.00 g of Fe are reacted with excess water, how many moles of each
product will be produced?
100.00gFe
= 1.7907 mol Fe
55.845g / mol
1.7907 mol Fe
1Fe3O4
= 0.59690 mol Fe3O4
3Fe
4H2
=
1.79 mol Fe 3Fe 2.3876 mol H2
1Fe3O4
= 0.59689 mol Fe3O4
3Fe
5. What is the weight of Fe3O4 that will be produced from the 100.00 g of Fe?
1.7907 mol Fe
3(55.845) + 4 (15.999) = 231.531 g/mol
0.59689 mol Fe3O4 (231.531) = 138.20 g Fe3O4
6. If 113.41 g of Fe3O4 is produced from 100.00 g of Fe, calculate the percent
yield for the reaction.
100.00 g of Fe should produce 138.21 g of Fe3O4
113.41
(100) = 8.2056 % yield
138.21
2Al + 3Cl2 ® 2AlCl3
7. How many grams of Al should you weigh out in order to produce 250.00 g of
AlCl3?
250.00
= 1.875 mol AlCl3
133.341
2Al
1.875
= 1.875 mol Al
2AlCl3
1.875 (26.982) = 50.588 g Al
8. How many moles of Cl2 gas will be needed to react with the quantity of Al
from problem 5?
3Cl2
1.875
= 2.8123 mol Cl2
2Al
9. If 50.589 g of Al actually produce 1.684 mol of AlCl3, determine the percent
yield for the reaction.
50.589 g of Al should produce 1.8749 mol of AlCl3
1.684
(100) = 89.82 % yield
1.8749
10. If you have 500.00 g of Al and 20.00 moles of Cl2 gas how many moles of AlCl3
can you produce?