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