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Question #67d21

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Feb 11, 2017

Approx. $3.5 \times 10^{3} \cdot g$ $3.5xx10^3*g$ .......

Explanation:

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

$= \frac{5000 \cdot g}{159.69 \cdot g \cdot m o l^{- 1}} = 31.3 \cdot m o l$ $=(5000*g)/(159.69*g*mol^-1)=31.3*mol$ .

But in each mole of $F e_{2} O_{3}$ $Fe_2O_3$ , clearly there are 2 moles of iron atoms, and 3 moles of oxygen atoms.

And thus $mass$ $"mass"$ $=$ $=$ $Molar mass \times number of moles$ $"Molar mass"xx"number of moles"$

$= 31.3 \cdot m o l \times 2 \times 55.85 \cdot g \cdot m o l^{- 1} =$ $=31.3*molxx2xx55.85*g*mol^-1=$ $? ? g$ $??g$

How many grams of oxygen.

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