# How many grams of iron are in 79.2 g of iron (Ill) oxide?

Feb 27, 2017

$\text{Mass of iron } \cong$ $\text{55"*"g}$

#### Explanation:

We calculate (i) a molar quantity of iron oxide:

$\frac{79.2 \cdot g}{159.69 \cdot g \cdot m o {l}^{-} 1} = 0.496 \cdot m o l$.

And then (ii) convert this molar quantity into a mass of iron:

$\text{Mass of iron} = 2 \times 0.496 \cdot m o l \times 55.85 \cdot g \cdot m o {l}^{-} 1 = 55.4 \cdot g$.

Why did I multiply the molar quantity by 2?