\(PTHH:2Zn+O_2\rightarrow2ZnO\)
\(n_{Zn}=\dfrac{52}{65}=0,8\left(mol\right)\)
\(n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Ta có:\(\dfrac{n_{Zn}}{2}:\dfrac{n_{O_2}}{1}=\dfrac{0,8}{2}< \dfrac{0,25}{1}\)⇒O2 pư hết
Chất A gồm Zn dư và ZnO.
Theo PTHH, ta có:\(n_{Zn}\left(pư\right)=2n_{O_2}=2.0,25=0,5\left(mol\right)\)
\(n_{Zn}\left(dư\right)=0,8-0,5=0,3\left(mol\right)\Rightarrow m_{Zn}trongA=0,3.65=19,5\left(g\right)\)
Theo PTHH ta có:\(n_{ZnO}=2n_{O_2}=2.0,25=0,5\left(mol\right)\Rightarrow m_{ZnO}=0,5.81=40,5\left(g\right)\)
Khối lượng chất rắn A là: 19,5+40,5=60(g)
\(\%Zn=\dfrac{19,5}{60}.100\%=32,5\%\)
\(\%ZnO=\dfrac{40,5}{60}.100\%=67,5\%\)