$解:生成氫氣的質(zhì)量為\mathrm{20g+50g-69.8g=0.2g}$
$則設(shè)樣品中鋅的質(zhì)量為x,生成硫酸鋅的質(zhì)量為y。$
$\ \text {Zn+H}_2\ \text {SO}_4\xlongequal[\ \ \ \ \ \ ]{}\ \text {ZnSO}_4+\ \text {H}_2↑$
65 161 2
x y 0.2g
$\mathrm{\ \frac{65}{x}=\frac{2}{0.2\mathrm{g}}=\frac{161}{y}\ \ 解得x=6.5g,y=16.1g\ \ }\ \ \ $
$鋅的質(zhì)量分?jǐn)?shù)為 \frac{6.5\mathrm{g}}{20\mathrm{g}}×100\%=32.5\%$
$答:鋅的質(zhì)量分?jǐn)?shù)為32.5\%。$