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58问答库 > 相同质量的铝、镁、铁、锌和足量的稀盐酸或稀硫酸充分反应，生成氢气的质量由多到少的质量为？

相同质量的铝、镁、铁、锌和足量的稀盐酸或稀硫酸充分反应，生成氢气的质量由多到少的质量为？

2025-04-03 22:56:57

推荐回答（1个）

回答（1）：

M(Al)=27；M(Mg)=24；M(Fe)=56；M(Zn)=65;顺序由多到少为：Al>Mg>Fe>Zn。原因是请计算产生1molH，所需要的金属量，则m(Al)=9g；m(Mg)=12；m(Fe)=28;m(Zn)=32.5g。

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