`n_{H_2}=\frac{0,896}{22,4}=0,04(mol)`
Cho $\begin{cases} Mg : x(mol)\\ Al: y(mol)\\\end{cases}$
`24x+27y=0,78g(1)`
BTe:
$\mathop{Mg}\limits^{0}\to \mathop{Mg}\limits^{+2}+2e$
$\mathop{Al}\limits^{0}\to \mathop{Al}\limits^{+3}+3e$
$\mathop{2H}\limits^{+}+2e\to \mathop{H_2}\limits^{0}$
`=> 2x+3y=0,08(mol)(2)`
`(1),(2)=>x=0,01(mol), y=0,02(mol)`
`a) ` `m_{Mg}=24.0,01=0,24g`
`b)` `n_{H_2SO_4}=n_{H_2}=0,04(mol)`
BTKL: `m=m_{\text{hh kim loại}}+m_{H_2SO_4}-m_{H_2}`
`=> m=0,78+0,04.98-0,04.2=4,62g`
`c)` `m_{\text{dd} H_2SO_4}=\frac{0,04.98}{9,8%}.100%=40g`
