Question

A chemist determined by measurement that 0.0450 moles of iron participated in a chemical reaction. Calculate the mass of iron that participated in the chemical reaction.

Answer 1

Answer:

The amount of Iron that participated in the chemical reaction was 2,51g

Explanation:

First: We look out for the atomic weight of  Iron, as we can see in the image below, its atomic weight is 55.845 g/ mol

Second: Now, we multiply the amoung of moles with the atomic weight in order to cancel units of moles :

$\text{Mass of Fe} = \text{0.0450 mol} \times \dfrac{\text{55.84 g Fe}}{\text{1 mol Fe}} = \textbf{2.51 g Fe}$

Answer 2

Answer:

2.51 g

Explanation:

The mass of 1 mol of iron is 55.84 g.

For 0.0450 mol,

$\text{Mass of Fe} = \text{0.0450 mol} \times \dfrac{\text{55.84 g Fe}}{\text{1 mol Fe}} = \textbf{2.51 g Fe}$