Question

Answer 1

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

            \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

            \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

Theo hệ số trên PT: \(n_{HCl}=2n_{H_2}=0,7\left(mol\right)\) \(\Rightarrow m_{HCl}=0,7\cdot36,5=25,55\left(g\right)\)

Bảo toàn khối lượng: \(m_{muối}=m_{KL}+m_{HCl}-m_{H_2}=39,65\left(g\right)\)