Đổi: \(19,2cm^3=19,2ml=0,0192l\)
Zn + H2SO4 → ZnSO4 + H2↑
\(n_{H_2}=\dfrac{0,0192}{24}=0,0008\left(mol\right)\)
a) Theo PT: \(n_{H_2SO_4}pư=n_{H_2}=0,0008\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}pư=0,0008\times98=0,0784\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}bđ=\dfrac{0,0784}{5\%}=1,568\left(g\right)\)
b) Theo PT: \(n_{ZnCl_2}=n_{H_2}=0,0008\left(mol\right)\)
\(\Rightarrow m=m_{ZnCl_2}=0,0008\times136=0,1088\left(g\right)\)
Zn + H2SO4 → ZnSO4 + H2
Đổi: \(19,2cm^3=19,2ml=0,0192l\)
\(n_{H_2}=\dfrac{0,0192}{24}=0,0008\left(mol\right)\)
a) Theo PT: \(n_{H_2SO_4}pư=n_{H_2}=0,0008\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}pư=0,0008\times98=0,0784\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}bđ=\dfrac{0,0784}{\left(100\%-5\%\right)}=0,0825\left(g\right)\)
b) Theo PT: \(n_{ZnSO_4}=n_{H_2}=0,0008\left(mol\right)\)
\(\Rightarrow m=m_{ZnSO_4}=0,0008\times161=0,1288\left(g\right)\)