\(n_{Zn}=\dfrac{9.6}{65}=\dfrac{48}{325}\left(mol\right)\)
a. \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b. Từ pthh suy ra: \(n_{HCl}=2n_{Zn}=2.\dfrac{48}{325}=\dfrac{96}{325}\left(mol\right)\)
=> \(CM_{HCl}=\dfrac{96:325}{0,4}=0,74\left(M\right)\)
c. dd B: \(ZnCl_2\)
\(ZnCl_2+2KOH\rightarrow2KCl+Zn\left(OH\right)_2\)
có: \(n_{Zn\left(OH\right)_2}=n_{ZnCl_2}=n_{Zn}=\dfrac{48}{325}\left(mol\right)\)
=> \(m_{kt}=\dfrac{48}{325}.99=14,62\left(g\right)\)
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