a) \(n_{H_2}:\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Gọi x, y lần lượt là số mol của \(Fe_3O_4,ZnO\)
\(Fe_3O_4+4H_2\rightarrow3Fe+4H_2O\)
1...................4............3............4(mol)
x..................4x.........3x...........4x(mol)
\(ZnO+H_2\rightarrow Zn+H_2O\)
1..............1...........1.........1(mol)
y..............y............y.........y(mol)
Ta có:\(\left\{{}\begin{matrix}232x+81y=19,7\\4x+y=0,3\end{matrix}\right.\)
=>x=0,05
=>y=0.1
\(m_{Fe_3O_4}:232.0,05=11,6\left(g\right)\)
\(m_{ZnO}:19,7-11,6=8,1\left(g\right)\)
b)\(m_{Fe}:56.0,15=8,4\left(g\right)\)
\(m_{Zn}:65.0,1=6,5\left(g\right)\)
c)\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
....1................1..................1............1(mol)
0,3................0,3................0,3.........0,3(mol)
\(m_{Mg}:0,3.24=7,2\left(g\right)\)
\(m_{H_2SO_4}:0,3.98+0,3.98.10\%=32.34\left(g\right)\)
