Đáp án:
\( {m_{CuS{O_4}.5{H_2}O}} = 62,5{\text{ gam}}\)
\( {m_{{H_2}O}} = 187,5{\text{ gam}}\)
Giải thích các bước giải:
Trong 250 gam dung dịch \(CuSO_4\) 16% chứa
\({m_{CuS{O_4}}} = 250.16\% = 40{\text{ gam}}\)
\( \to {n_{CuS{O_4}}} = \frac{{40}}{{64 + 96}} = 0,25{\text{ mol = }}{{\text{n}}_{CuS{O_4}.5{H_2}O}}\)
\( \to {m_{CuS{O_4}.5{H_2}O}} = 0,25.(64 + 96 + 18.5) = 62,5{\text{ gam}}\)
\( \to {m_{{H_2}O}} = 250 - 62,5 = 187,5{\text{ gam}}\)
`mct_{CuSO_4. 5H_2O}=62,5(g)`
`m_{H_2O}=187,5(g)`
`n_{CuSO_4}=\frac{250.16%}{160.100%}=0,25(mol)`
Bảo toàn nguyên tố `Cu`
`n_{CuSO_4. 5H_2O}=n_{CuSO_4}=0,25(mol)`
` \to mct_{CuSO_4. 5H_2O}=0,25.(160+5.18)=62,5(g)`
`m_{H_2O}=mdd_{CuSO_4}-mct_{CuSO_4. 5H_2O}=250-62,5=187,5(g)`
Giups mik vs trả lời bằng hình chụp càng tốt
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