解:(1)設(shè)參與反應(yīng)的CaCl
2質(zhì)量為x,K
2CO
3的質(zhì)量為y,生成KCl的質(zhì)量為z,
CaCl
2+K
2CO
3=CaCO
3↓+2KCl
111 138 100 149
x y 10g z
∴
,
,
,
解之得:x=
=11.1g;
y=
=13.8g;
z=
=14.9g;
(2)CaCl
2溶液中溶質(zhì)的質(zhì)量分數(shù)=
×100%=11.1%;
(3)反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分數(shù)為:
×100%=
×100%≈15.3%.
答:(1)原混合物中含碳酸鉀的質(zhì)量為13.8g;
(2)CaCl
2溶液中溶質(zhì)的質(zhì)量分數(shù)為11.1%;
(3)反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分數(shù)為15.3%.
分析:氯化鉀不能與CaCl
2發(fā)生反應(yīng),故混合物中只有碳酸鉀與CaCl
2反應(yīng);
(1)利用碳酸鉀與CaCl
2反應(yīng)的化學(xué)方程式和生成沉淀的質(zhì)量,列出比例式,就可計算出參與反應(yīng)的CaCl
2質(zhì)量和K
2CO
3的質(zhì)量,以及生成KCl的質(zhì)量;
(2)根據(jù)“溶質(zhì)質(zhì)量分數(shù)=
×100%”就可計算出CaCl
2溶液中溶質(zhì)的質(zhì)量分數(shù);
(3)反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分數(shù)=
×100%.
點評:本題主要考查學(xué)生利用化學(xué)方程式和溶質(zhì)質(zhì)量分數(shù)公式進行計算的能力.計算時要注意反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量包括混合物中氯化鉀的質(zhì)量和反應(yīng)生成的氯化鉀的質(zhì)量;反應(yīng)后所得溶液質(zhì)量需減去生成沉淀的質(zhì)量.