a) \(H_2+ZnO\rightarrow Zn+H_2O\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{4,48}{22,4}=0,2mol\)
\(n_{ZnO}=\dfrac{m_{ZnO}}{M_{ZnO}}=\dfrac{24,3}{81}=0,3mol\)
Ta có: \(\dfrac{n_{H_2}}{1}=\dfrac{0,2}{1}\)
\(\dfrac{n_{ZnO}}{1}=\dfrac{0,3}{1}\)
\(\Rightarrow\dfrac{n_{H_2}}{1}< \dfrac{n_{ZnO}}{1}\)
Vậy ZnO dư
\(n_{ZnO\text{pứ}}=\dfrac{0,2.1}{1}=0,2mol\)
\(\Rightarrow n_{ZnO\text{dư}}=n_{ZnO}-n_{ZnO\text{pứ}}=0,3-0,2=0,1mol\)
Khối lượng ZnO dư:
\(m_{ZnO\text{dư}}=n_{ZnO\text{dư}}.M_{ZnO}=0,1.81=8,1g\)
b) Theo PTHH: \(n_{Zn}=\dfrac{0,2.1}{1}=0,2mol\)
Khối lượng chất rắn thu được:
\(m_{Zn}=n_{Zn}.M_{Zn}=0,2.65=13g\)