# Question 67d21

Feb 11, 2017

Approx. $3.5 \times {10}^{3} \cdot g$.......

#### Explanation:

$\text{Number of moles of ferric oxide}$ $=$ $\text{Mass"/"Molar mass}$

$= \frac{5000 \cdot g}{159.69 \cdot g \cdot m o {l}^{-} 1} = 31.3 \cdot m o l$.

But in each mole of $F {e}_{2} {O}_{3}$, clearly there are 2 moles of iron atoms, and 3 moles of oxygen atoms.

And thus $\text{mass}$ $=$ $\text{Molar mass"xx"number of moles}$

$= 31.3 \cdot m o l \times 2 \times 55.85 \cdot g \cdot m o {l}^{-} 1 =$ ??g#

How many grams of oxygen.