# 11

1. The reaction is as follows:

Mg(s) + H2SO4(aq) $\to$ MgSO​4(aq) + H2(g)

1 mole of Mg gives 1 mole H2

2. Moles of H2 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  of H2

3.  0.012 moles of H2=0.012 moles of Mg
mass of Mg =
=0.319

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

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