a, Ta có: \(n_{H_2}=0,5\left(mol\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
___0,5____________________0,5 (mol)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(\Rightarrow m_{Mg}=0,5.24=12\left(g\right)\)
\(m_{MgO}=20-12=8\left(g\right)\)
b, Ta có: \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,7\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,7.98=68,6\left(g\right)\)
\(\Rightarrow C\%_{ddH_2SO_4}=\dfrac{68,6}{200}.100\%=34,3\%\)
c, Theo PT: \(n_{MgSO_4}=n_{Mg}+n_{MgO}=0,7\left(mol\right)\)
Có: m dd sau pư = 20 + 200 - 0,5.2 = 219 (g)
\(\Rightarrow C\%_{MgSO_4}=\dfrac{0,7.120}{219}.100\%\approx38,36\%\)
