Stoich CW 2 Ans
C3H8 + 5O 2 ® 3CO 2 + 4H 2O
1. Determine the moles of propane, C3H8, needed to produce 23.1 g of water.
a. Convert g of H2O to mol of H2O
23.1
= 1.282279 mol of H2O
18.0148
(Remember not to round till the end of a problem)
b. Convert mol H2O to mol of C3H8
1C H
1.282279 3 8 = 0.321 mol of C3H8
4H 2O
1. Calculate the mass of 1.84 x 10-5 moles of water.
1.84 x 10-5 (18.0148) = 3.31 x 10-4 g of H2O
2. How many moles of CO2 are will be produced upon the reaction of 1.745
moles of O2
3CO2
1.745
= 1.047 mol CO2
5O2
3. 0.148 moles of propane will produce what mass of carbon dioxide.
3CO2
0.148
= 0.444 mol CO2
1C3 H 8
0.444 (44.009) = 19.5 g of CO2
4. How many grams of O2 would you need to react with 125.5g of C3H8?
125.5
= 2.846 mol C3 H 8
44.096
5O2
2.846 mol C3 H 8
= 14.23 mol O2
1C3 H 8
14.23 mol O2 · 31.998 = 455.3 g O2
3Fe + 4H 2O ® Fe3O4 + 4H 2
5. If 100.00 g of Fe are reacted with excess water, how many moles of each
product will be produced?
100.00 / 55.845 = 1.7907 mol Fe
1Fe3O4
1.7907
= 0.59690 mol Fe3O4
3Fe
4H2
1.7907
= 2.3876 mol H2
3Fe
6. What is the mass of Fe3O4 that will be produced from 125.0 g of water?
125.0 / 18.015 = 6.939 mol H2O
1Fe3O4
6.939
= 1.735 mol Fe3O4
4H 2O
1. 735 (231.53) = 401.7 g Fe3O4
7. Calculate the moles of iron necessary to produce 8.45 moles of H2.
3Fe
8.45
= 6.34 mol of Fe
4H 2
2Al + 3Cl2 ® 2AlCl3
8. How many grams of Al should you weigh out in order to produce 250.00 g of
AlCl3?
250.00 / 133.34 = 1.8749 mol AlCl3
2Al
1.8749
=1.8749 mol Al
2AlCl3
1.8749 (26.982) = 50.589 g of Al
9. How many moles of Cl2 gas will be needed to react with the quantity of Al
from problem 4?
3Cl2
1.8749
= 2.8123 mol Cl2
2Al
10. If you have 500.00 g of Al how many grams of AlCl3 can you produce?
500.00 / 26.982 = 18.531 mol Al
2AlCl3
18.531 2Al = 18.531 mol AlCl3
18.531 (133.34) = 2470.9 g of AlCl3
4NH 3 + 5O2 ® 4NO + 6H 2 0
11. How many grams of each reactant are required to produce 4.250 moles of
NO?
4NH 3
= 4.250 moles NH 3
4.250 moles of NO 4NO
4.250 (17.031) = 72.38 g of NH3
5O2
4.250 moles of NO
= 5.313 moles O2
4NO
5.313 (31.998) = 170.0 g of O2
12. Determine the mass of water produced from reacting 250.0 g of NH3
250.0 / 17.031 = 14.68 mol NH3
6H 2O
14.68
= 22.02 mol H2O
4 NH 3
22.02 (18.015) = 396.7 g H2O