Question

For the following reaction, 37.3 grams of iron are allowed to react with 20.0 grams of oxygen gas. What is the maximum amount of iron(III) oxide that can be formed? What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete?

(A) Maximum iron(III) oxide formed: 57.2 g; Limiting reagent: Fe; Excess reagent remaining: 12.8 g
(B) Maximum iron(III) oxide formed: 57.2 g; Limiting reagent: O2; Excess reagent remaining: 12.8 g
(C) Maximum iron(III) oxide formed: 49.6 g; Limiting reagent: Fe; Excess reagent remaining: 20.0 g
(D) Maximum iron(III) oxide formed: 49.6 g; Limiting reagent: O2; Excess reagent remaining: 37.3 g

Answer 1

To find the maximum amount of iron(III) oxide formed, the limiting reagent, and the excess reagent remaining, one must use stoichiometry with the balanced chemical equation, starting masses of reactants, and molar masses of the compounds involved.

Step-by-step explanation:

The question requires us to find the maximum amount of iron(III) oxide formed from the reaction between iron and oxygen, identify the limiting reagent, and determine the mass of excess reagent remaining after the reaction is complete. We need to use stoichiometry to compute these values based on the starting masses of iron (Fe) and oxygen (O2).

To solve this problem, we must use the balanced chemical equation for the formation of iron(III) oxide (Fe2O3):

1. 4 Fe + 3 O2 → 2 Fe2O3

By converting the masses of reactants to moles, comparing the mole ratios from the balanced equation, and determining which reactant will run out first (thus being the limiting reagent), we can then calculate the mass of the Fe2O3 produced.

After identifying the limiting reagent, we calculate the mass of the remaining excess reagent by determining how much of it was used in the reaction and subtracting this from the initial mass provided.