Đáp án:
`CuO + H_2SO_4 → CuSO_4 + 2H_2O`
`n_{CuO} = 0,8: 80 = 0,01mol.`
`n_{H_2SO_4} = 0,03mol.`
Theo phương trình hóa học:
Số mol `H_2SO_4 dư = 0,03 - 0,01 = 0,02 mol.`
Dung dịch thu được sau phản ứng có `0,02` mol `H_2SO_4` và `0,01` mol `CuSO_4.`
Giải thích các bước giải:
n CuO=0,8\80=0,01 mol
nH2SO4=0,03 mol
CuO+ H2SO4->CuSO4+H2O
0,01-----0,01-------0,01 mol
=>n H2SO4 dư = 0,03-0,01 =0,02 mol
=> n CuSO4 = 0,01 mol
vậy chất còn lại là H2SO4 dư và CuSO4
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