Question

Answer 1

a) \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right);n_{H_2SO_4}=\dfrac{14,7.20\%}{98}=0,03\left(mol\right)\)

Lập tỉ lệ : \(\dfrac{0,02}{1}< \dfrac{0,03}{1}\)

=> Sau phản ứng H2SO4 dư

\(m_{H_2}=0,02.22,4=0,448\left(l\right)\)

b) \(n_{H_2SO_4\left(dư\right)}=0,03-0,02=0,01\left(mol\right)\)

\(n_{ZnSO_4}=n_{Zn}=0,02\left(mol\right)\)

=> \(m_{H_2SO_4}=0,01.98=0,98\left(g\right)\)

\(m_{ZnSO_4}=0,02.161=3,22\left(g\right)\)