Mỗi phần có $5g$ $X$
- P1:
BTKL: $m_{O_2}=21-5=16g$
$\to n_{O_2}=\dfrac{16}{32}=0,5(mol)$
$2Mg+O_2\xrightarrow{{t^o}} 2MgO$
$4Al+3O_2\xrightarrow{{t^o}} 2Al_2O_3$
$2Zn+O_2\xrightarrow{{t^o}} 2ZnO$
$\to n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{3}{4}n_{Al}+\dfrac{1}{2}n_{Zn}$
$\to 4n_{O_2}=2n_{Mg}+3n_{Al}+2n_{Zn}$
- P2:
$Mg+4HNO_3\to Mg(NO_3)_2+2NO_2+2H_2O$
$Al+6HNO_3\to Al(NO_3)_3+3NO_2+3H_2O$
$Zn+4HNO_3\to Zn(NO_3)_2+2NO_2+2H_2O$
$\to n_{NO_2}=2n_{Mg}+3n_{Al}+2n_{Zn}$
$\to n_{NO_2}=4n_{O_2}=0,5.4=2(mol)$
Vậy $V=44,8l$