Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
____0,2____0,2______0,2_____0,2 (mol)
a, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
b, \(m_{H_2SO_4}=0,2.98=19,6\left(g\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{19,6}{200}.100\%=9,8\%\)
c, Có: m dd sau pư = 4,8 + 200 - 0,2.2 = 204,4 (g)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,2.120}{204,4}.100\%\approx11,74\%\)