Answer:
14.25 g of Fe2O3
33.26 g of Fe2O3
0.045 moles of excess reagent.
Explanation:
The equation for the reaction is shown below, remember that noting the correct chemical reaction equation is the first step towards solving the problem.
4Fe (s) + 3O2 (g) ==> 2 Fe2O3 (s)
Since the oxygen is abundant in air, oxygen is the reactant in excess while the iron is the limiting reactant. Remember that only a thin iron metal burns according to the question.
To double check our assumption above, the reactant that gives the least mass of product is the limiting reactant.
Using iron;
If 224 g of iron yields 319.38 g of Fe2O3
10g of iron will yield 10 × 319.38/224 = 14.25 g of Fe2O3
Using oxygen
96 g of oxygen yields 319.38 g of Fe2O3
10 g of oxygen will yield 10 × 319.38/96 = 33.26 g of Fe2O3
Since iron is the limiting reagent, number of moles in 10g of iron = 10g/56gmol-1 = 0.18 moles
From the reaction equation
4 moles of iron reacts with 3 moles of oxygen
0.18 moles of iron will react with 0.18 ×3/4 = 0.135 moles of oxygen
Amount of excess reagent remaining; 0.18- 0.135 = 0.045 moles of excess reagent.
