解：根據(jù)質(zhì)量守恒定律，反應(yīng)生成氫氣的質(zhì)量為?$20.8\mathrm {g}+179.6\mathrm {g}-200\mathrm {g}=0.4\mathrm {g}$?
設(shè)參加反應(yīng)的鋅的質(zhì)量為?$x，$?反應(yīng)生成硫酸鋅的質(zhì)量為?$y。$?
?$\mathrm {Zn}+\mathrm {H}_2\mathrm {SO}_4\xlongequal[ ]{ }\mathrm {ZnSO}_4+\mathrm {H}_2↑$?
65 161 2
? $x$? y 0.4g
?$\frac {65 }{x }=\frac {161 }{y }=\frac {2 }{0.4\mathrm {g} }，$?解得?$x=13\mathrm {g}，$??$y=32.2\mathrm {g}$?
?$(1\mathrm {)}$?混合物中鋅的質(zhì)量分?jǐn)?shù)為?$\frac {13 \mathrm {g}}{20.8 \mathrm {g}}×100％=62.5％ $?
?$(2\mathrm {)}$?反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分?jǐn)?shù)：
?$\frac {32.2\mathrm {g}+(20.8\mathrm {g}-13\mathrm {g)}}{200\mathrm {g}}×100\%=20\%$
?
答：?$(1\mathrm {)}$?混合物中鋅的質(zhì)量分?jǐn)?shù)為?$62.5\%；$?
?$(2\mathrm {)}$?反應(yīng)后所得溶液的溶質(zhì)質(zhì)量分?jǐn)?shù)為?$20\%。$?
