Mỗi phần có $11,05g$ $X$ 

- P1:

BTKL: $m_{O_2}=18,25-11,05=7,2g$

$\to n_{O_2}=\dfrac{7,2}{32}=0,225(mol)$

$2Mg+O_2\xrightarrow{{t^o}} 2MgO$ 

$4Al+3O_2\xrightarrow{{t^o}} 2Al_2O_3$

$2Zn+O_2\xrightarrow{{t^o}} 2ZnO$

$\to n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Zn}$

$\to 4n_{O_2}=2n_{Mg}+3n_{Al}+2n_{Zn}$ 

- P2:

$Mg+H_2SO_4\to MgSO_4+H_2$

$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$

$Zn+H_2SO_4\to ZnSO_4+H_2$

$\to n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}+n_{Zn}$

$\to 2n_{H_2}=2n_{Mg}+3n_{Al}+2n_{Zn}$

$\to 2n_{H_2}=4n_{O_2}=0,225.4=0,9(mol)$

$\to n_{H_2}=0,45(mol)$

$\to V=0,45.22,4=10,08l$

Theo PTHH, $n_{H_2SO_4}=n_{H_2}=0,45(mol)$

BTKL:

$m=11,05+0,45.98-0,45.2=54,25g$