a) \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
0,1<--0,4-------->0,3
=> \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)
\(\%m_{Fe_3O_4}=\dfrac{23,2}{31,2}.100\%=74,359\%\)
\(\%m_{MgO}=100\%-74,359\%=25,641\%\)
b)
\(n_{MgO}=\dfrac{31,2-23,2}{40}=0,2\left(mol\right)\)
PTHH: Fe + H2SO4 --> FeSO4 + H2
0,3-->0,3
MgO + H2SO4 --> MgSO4 + H2O
0,2--->0,2
=> \(m_{H_2SO_4}=\left(0,3+0,2\right).98=49\left(g\right)\)
=> \(C\%=\dfrac{49}{300}.100\%=16,33\%\)
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