a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
b, Ta có: \(n_{Zn}=\dfrac{1,95}{65}=0,03\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{1,47}{98}=0,015\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,015}{1}\), ta được Zn dư.
Theo PT: \(n_{Zn\left(pư\right)}=n_{H_2SO_4}=0,015\left(mol\right)\Rightarrow n_{Zn\left(dư\right)}=0,03-0,015=0,015\left(mol\right)\)
\(\Rightarrow m_{Zn\left(dư\right)}=0,015.65=0,975\left(g\right)\)
c, \(n_{H_2}=n_{H_2SO_4}=0,015\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,015.22,4=0,336\left(l\right)\)