Tính khối lượng MgCl2 tạo thành khi hòa tan 7,2g Mg bằng HCl dư?

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{7,2}{24}=0,3\left(mol\right)\)

Theo PTHH ,có :

\(n_{MgCl_2}=n_{H_2}=n_{Mg}=0,3\left(mol\right)\)

\(m_{MgCl_2}=n.M=0,3.\left(24+2.35,5\right)=28,5\left(g\right)\)

\(V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)