a, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,2 0,4 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b,\(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{200}=7,3\%\)
c, mdd sau pứ = 4,8+200-0,2.2 = 204,4 (g)
\(C\%_{ddMgCl_2}=\dfrac{0,2.95.100\%}{204,4}=9,3\%\)