B. If 10.0 g iron reacts with oxygen, how many grams of iron (III) oxide should be
produced?

14.29g

Explanation:

First, let us write a balanced equation for the reaction. This is illustrated below below:

4Fe + 3O2 —> 2Fe2O3

Next, let us calculate the mass of Fe that reacted and the mass of Fe2O3 produced from the balanced equation. This is shown below:

Molar Mass of Fe = 56g/mol

Mass of Fe from the balanced equation = 4 x 56 = 224g

Molar Mass of Fe2O3 = (56x2) + (16x3) = 112 + 48 = 160g/mol

Mass of Fe2O3 from the balanced equation = 2 x 160 = 320g

From the balanced equation,

224g of Fe produced 320g of Fe2O3.

Therefore, 10g of Fe will produce = (10x320)/224 = 14.29g of Fe2O3

From the calculations made above, 10g iron(Fe) produced 14.29g of iron(iii) oxide (Fe2O3)