# 11

Mg(s) + H

_{2}SO

_{4}(aq) $\to $ MgSO

_{4}(aq) + H

_{2}(g)

1 mole of Mg gives 1 mole H

_{2}

2. Moles of H

_{2}obtained:

24000 cc is the volume of 1 mole of gas at STP

1 cc is the volume of $\frac{1}{24000}$ mole of gas

298.6 cc is $\frac{1}{24000}\times 298.6=0.012moles$ of H

_{2}

3. 0.012 moles of H

_{2}=0.012 moles of Mg

mass of Mg = $numberofmoles\times molarmass\phantom{\rule{0ex}{0ex}}=0.012\times 24.3$

=0.319

4. Percent purity = $\frac{0.319}{1}\times 100=31.99\%$

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