将5.6g铁放入足量的稀硫酸中充分反应，得到80mL密度为1.25g/mL的溶液_爱下问搜题

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问题问答题

将5.6g铁放入足量的稀硫酸中充分反应，得到80mL密度为1.25g/mL的溶液．试计算：

（1）生成标准状况下的H₂多少升？

（2）所得溶液中硫酸亚铁的物质的量浓度．

答案

（1）硫酸足量，5.6g铁完全反应，令生成生成标准状况下的H₂的体积为V，则：

Fe+H₂SO₄=FeSO₄+H₂

56g 22.4L

5.6g V

所以V=

5.6g

56g

×22.4L=2.24L，

答：5.6g铁完全反应生成标准状况下H₂的体积为2.24L；

（2）5.6g铁完全反应，根据铁元素守恒可知，溶液中n（FeSO₄）=n（Fe）=

5.6g

56g/mol

=0.1mol，溶液中c（FeSO₄）=

0.1mol

0.08L

=1.25mol/L，

答：溶液中硫酸亚铁的物质的量浓度为1.25mol/L．

完形填空

It Seemed Like a Good Idea at the Time

I really only meant it as a joke. I wasn't trying to steal anything honestly.
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