(2011?黃岡模擬)將13g Cu粉和Zn粉的混合物放入燒杯中,向其中加入154.7g稀硫酸,恰好完全反應(yīng)后,得到0.2g氫氣.求:
(1)混合物中Cu粉的質(zhì)量分數(shù).
(2)反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分數(shù).
【答案】
分析:(1)混合物中Cu粉的質(zhì)量分數(shù)=
×100%,根據(jù)鋅與硫酸恰好完全反應(yīng)放出氫氣的質(zhì)量計算出混合物中所含鋅的質(zhì)量;
(2)反應(yīng)后所得溶液為硫酸鋅溶液,其中溶質(zhì)的質(zhì)量分數(shù)
×100%,硫酸鋅的質(zhì)量可根據(jù)反應(yīng)的化學(xué)方程式由放出氫氣質(zhì)量求得,而反應(yīng)后溶液的質(zhì)量則可根據(jù)質(zhì)量守恒定律求得.
解答:解:設(shè)恰好完全反應(yīng)時,消耗鋅的質(zhì)量為x,生成硫酸鋅的質(zhì)量為y
Zn+H
2SO
4═ZnSO
4+H
2↑
65 161 2
x y 0.2g
=
x=6.5g
=
y=16.1g
(1)混合物中Cu粉的質(zhì)量分數(shù)=
×100%=50%
(2)反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分數(shù)=
×100%=10%
答:(1)混合物中Cu粉的質(zhì)量分數(shù)為50%;(2)反應(yīng)后所得溶液中溶質(zhì)的質(zhì)量分數(shù)為10%.
(其它合理解法參照給分).
點評:根據(jù)反應(yīng)的化學(xué)方程式可以表示反應(yīng)中各物質(zhì)的質(zhì)量比,由反應(yīng)中某物質(zhì)的質(zhì)量可計算出反應(yīng)中其它物質(zhì)的質(zhì)量.