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

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

Step-by-step 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} * \frac{\text{55.84 g Fe}}{\text{1 mol Fe}} = \textbf{2.51 g Fe}`

Answer 2

2.51 g

Step-by-step explanation:

The mass of 1 mol of iron is 55.84 g.

For 0.0450 mol,

`\text{Mass of Fe} = \text{0.0450 mol} * \frac{\text{55.84 g Fe}}{\text{1 mol Fe}} = \textbf{2.51 g Fe}`