PROBLEM 35P
For the reaction shown, calculate how many grams of each product form when the given amount of each reactant completely reacts to form products. Assume there is more than enough of the other reactant.
2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(l)
(a) 4.7 g Al
(b) 4.7 g Fe2O3
Solution 35P
Here, we are going to calculate the mass of product formed when the given amount of each reactant completely reacts.
Step 1:
a)
The given reaction is:
2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(l)
Given, mass of Al = 4.7 g
Molar mass of Al = 26.98 g/mol
Therefore, number of moles of Al = given mass / molar mass
= = 0.17 mol
From the above reaction, it is seen that 2 moles of Al produces 1 mole of Al2O3 and 2 moles of Fe.
Therefore, 0.17 mol of Al will produce (½ x 0.17 = 0.085) mol of Al2O3 and 0.17 mol of Fe.
Step 2:
Now, 1 mol Al2O3 = molar mass of Al2O3
Therefore, 0.085 mol of Al2O3 = 0.085 x molar mass of Al2O3
= 0.085 x 101.96 g/mol
= 8.67 g
Again, 1 mol Fe = molar mass of Fe
Therefore, 0.17 mol of Fe = 0.17 x molar mass of Fe
= 0.17 x 55.845 g/mol
= 9.49 g
Thus 4.7 g Al produces 8.67 g Al2O3 and 9.49 g Fe.
Step3:
b)
Given, mass of Fe2O3 = 4.7 g
Molar mass of Fe2O3 = 159.69 g/mol
Therefore, number of moles of Fe2O3 = given mass / molar mass
= = 0.029 mol
From the above reaction, it is seen that 1 mole of Fe2O3 produces 1 mole of Al2O3 and 2 moles of Fe.
Therefore, 0.029 mol of Fe2O3 will produce 0.029 mol of Al2O3 and (0.029 x 2 = 0.058) mol of Fe.
Step 4:
Now, 1 mol Al2O3 = molar mass of Al2O3
Therefore, 0.029 mol of Al2O3 = 0.029 x molar mass of Al2O3
= 0.029 x 101.96 g/mol
= 2.96 g
Again, 1 mol Fe = molar mass of Fe
Therefore, 0.058 mol of Fe = 0.058 x molar mass of Fe
= 0.058 x 55.845 g/mol
= 3.24 g
Thus 4.7 g Fe2O3 produces 2.96 g Al2O3 and 3.24 g Fe.
