Cho một lượng mạt sắt dư vào 50ml dd HCl. Phản ứng xong, thu được 3,36 lít khí (đktc) a. Tính khối lượng mạt sắt tham gia dư b. Tìm nồng độ mol của dd HCl đã dùng

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Đáp án:

PTHH : `Fe + 2HCl -> FeCl_2 + H_2`

`a)`

`n_{H_2} = (3,36)/(22,4) = 0,15` `mol`

`n_{Fe} = n_{H_2} = 0,15` `mol`

`m_{Fe} = 0,15 . 56 = 8,4` `gam`

`b)`

`50ml = 0,05l`

`n_{HCl} = 2 . n_{H_2} = 0,3` `mol`

`C_{M_(HCl)} = (0,3)/(0,05) = 6` `M`

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