Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
____0,1______0,2______0,1____0,1 (mol)
a, mMg = 0,1.24 = 2,4 (g)
b, \(m_{ddHCl}=\dfrac{0,2.36,5}{14,6\%}=50\left(g\right)\)
c, mMgCl2 = 0,1.95 = 9,5 (g)