51问答网 > 黄铜是铜和锌的合金。向一定质量的黄铜样品中加入154.7g稀硫酸，恰好完全反应，产生0.2g氢气，

黄铜是铜和锌的合金。向一定质量的黄铜样品中加入154.7g稀硫酸，恰好完全反应，产生0.2g氢气，

求反应后所得溶液中溶质的质量分数
2025-03-26 23:35:22
推荐回答（1个）
回答1：

设参加反应的锌的质量为x，生成硫酸锌的质量为y
Zn+H2SO4=ZnSO4+H2↑
65 161 2
x y 0.2g

65
x
=
2
0.2g
x=6.5g

161
y
=
2
0.2g
y=16.1g
反应后所得溶液中溶质的质量分数=
16.1g
6.5g+154.7g-0.2g
×100%=10%
答：反应后所得溶液中溶质的质量分数为10%；

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