a) Mg + 2HCl --> MgCl2 + H2
b) Theo ĐLBTKL: mMg + mHCl = mMgCl2 + mH2 (1)
c)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=>m_{H_2}=0,5.2=1\left(g\right)\)
(1) => mMgCl2 = 12+36,5-1 = 47,5(g)
a: \(Mg+2HCl->MgCl_2+H_2\)
b: \(n_H=\dfrac{11.2}{22.4}=0.5\)
\(\Leftrightarrow m_H=M_H\cdot n_H=0.5\cdot2=1\left(g\right)\)