Ta thấy khi A + HCl giải phóng H2 → A có Zn dư
\(n_{H_2}=\frac{16,8}{22,4}=0,75\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
(mol)___0,75____________0,75____0,75_
\(m_{Zn}=65.0,75=48,75\left(g\right)\Rightarrow m_{ZnO}=146,25-48,75=97,5\left(g\right)\Rightarrow n_{ZnO}=\frac{97,5}{81}=1,2\left(mol\right)\)
\(PTHH:2Zn+O_2\underrightarrow{t^o}2ZnO\)
(mol)____1,2__________1,2__
\(m_{Zn\cdot pu}=1,2.65=78\left(g\right)\)
\(PTHH:ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
(mol)_____1,2_____________1,2__________
\(m_{ZnCl_2}=136.\left(1,2+0,75\right)=265,2\left(g\right)\)