Ta có: \(n_{Mg}=\dfrac{96}{24}=4\left(g\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
___4_______________4_______4 (mol)
a, \(V_{H_2}=4.22,4=89,6\left(g\right)\)
b, Ta có: m dd sau pư = 96 + 500 - 4.2 = 588 (g)
\(C\%_{MgSO_4}=\dfrac{4.120}{588}.100\%\approx81,63\%\)