设可制的氢气质量为x.
Zn+2HCl=ZnCl2+H2
73 2
73g*10% x
73/7.3g=2/x
x=0.2g
答:可制得氢气质量为0.2g.
溶质质量分数为%10的盐酸73g与足量的锌粒反应可制得多少克氢气
溶质质量分数为%10的盐酸73g与足量的锌粒反应可制得多少克氢气
化学人气:717 ℃时间:2019-11-01 14:36:56
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