Answer:
Option A. 1.02g of Fe.
Explanation:
Step 1:
The balanced equation for the reaction.
3Mg + Fe2(SO4)3 —> 3MgSO4 + 2Fe
Step 2:
Determination of the mass Fe2(SO4)3 that reacted and the mass of Fe produced from the balanced equation. This is illustrated below:
Molar Mass of of Fe2(SO4)3 = (2x56) + 3[32 + (16x4)] = 112 + 3[32 + 64] = 112 + 3[96] = 400g/mol
Mass of Fe2(SO4)3 from the balanced equation = 1 x 400 = 400g
Molar Mass of Fe = 56g/mol
Mass of Fe from the balanced equation = 2 x 56 = 112g
From the balanced equation above,
400g of Fe2(SO4)3 reacted to produce 112g Fe.
Step 3:
Determination of the mass of Fe produced by reacting 3.65g of Fe2(SO4)3.
This is illustrated below:
From the balanced equation above,
400g of Fe2(SO4)3 reacted to produce 112g Fe.
Therefore, 3.65g of Fe2(SO4)3 will react to produce = (3.65 x 112)/400 = 1.02g of Fe.
Therefore, 1.02g of Fe is produced from 3.65g of Fe2(SO4)3.
