將25g石灰石(主要成分是CaCO3)放入73g稀鹽酸中(石灰石中的雜質(zhì)不跟鹽酸反應(yīng),也不溶于水)恰好完全反應(yīng),生成8.8g二氧化碳氣體.試求:
(1)石灰石中碳酸鈣的質(zhì)量分數(shù)為______;
(2)稀鹽酸中溶質(zhì)的質(zhì)量分數(shù)為______:
(3反應(yīng)后溶液中溶質(zhì)的質(zhì)量分數(shù)為______.
解:設(shè)石灰石中碳酸鈣的質(zhì)量為x,稀鹽酸中溶質(zhì)的質(zhì)量為y,生成氯化鈣的質(zhì)量為z.
CaCO
3+2HCl=CaCl
2+H
2O+CO
2↑
100 73 111 44
x y z 8.8g
解得:x=20g y=14.6g z=22.2g
(1)石灰石中碳酸鈣的質(zhì)量分數(shù)為:
;
(2)稀鹽酸中溶質(zhì)的質(zhì)量分數(shù)為:
=20%
(3反應(yīng)后溶液中溶質(zhì)的質(zhì)量分數(shù)為:
=26.4%.
答:(1)石灰石中碳酸鈣的質(zhì)量分數(shù)為80%;(2)稀鹽酸中溶質(zhì)的質(zhì)量分數(shù)為20%;3)反應(yīng)后溶液中溶質(zhì)的質(zhì)量分數(shù)為26.4%.
分析:根據(jù)碳酸鈣與鹽酸反應(yīng)的化學(xué)方程式,由生成的二氧化碳的質(zhì)量求出碳酸鈣、鹽酸溶液中的溶質(zhì)、生成的氯化鈣的質(zhì)量,在計算出石灰石中碳酸鈣的質(zhì)量分數(shù)、稀鹽酸中溶質(zhì)的質(zhì)量分數(shù)和生成氯化鈣的質(zhì)量,進而計算出反應(yīng)后溶液中溶質(zhì)的質(zhì)量分數(shù).
點評:本題主要考查含雜質(zhì)物質(zhì)的化學(xué)方程式計算和溶質(zhì)質(zhì)量分數(shù)的計算,在計算時,要正確地根據(jù)質(zhì)量守恒定律計算反應(yīng)的溶液的質(zhì)量,難度較大.