a, Ta có: \(n_{H_2}=0,1\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
Theo PT: nZn = nH2 = 0,1 (mol)
⇒ mZn = 0,1.65 = 6,5 (g), mZnO = 14,6 - 6,5 = 8,1 (g)
b, Ta có: \(m_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
Theo PT: \(n_{ZnCl_2}=n_{Zn}+n_{ZnO}=0,2\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
c, Theo PT: \(n_{HCl}=2n_{Zn}+2n_{ZnO}=0,4\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{36\%}=\dfrac{365}{9}\left(g\right)\)
\(\Rightarrow V_{ddHCl}=\dfrac{\dfrac{365}{9}}{1,18}\approx34,37\left(g\right)\)