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16.7%

10%

35.7%

7.1%

$\text {H}_2\text {SO}_4$

9.8

$\text {H}_2\text {O}$

90.2

$\mathrm{ZnSO_4}$

16.1

15.1%

$\text {H}_2\text {O} $

90.2

$\text {CaCO}_3+2\text {HCl}\xlongequal[\ \ \ \ \ \ ]{\ \ \ \ \ \ }$

$\text {CaCl}_2+\text {H}_2\text {O}+\text {CO}_2↑$

$\text {CaCl}_2$

HCl

m(稀鹽酸)+m(大理石)-m(雜質(zhì))-m(二氧化碳)

D

B

D

106.3

$(2)解：反應(yīng)后溶液溶質(zhì)的質(zhì)量分?jǐn)?shù)：\frac {15.2\text{g}}{(200+5.6-0.2)\text{g}}×100\%=7.4\%$

$答：反應(yīng)后溶液溶質(zhì)質(zhì)量分?jǐn)?shù)為7.4\%。$

(更多請(qǐng)點(diǎn)擊查看作業(yè)精靈詳解)

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