-Gọi $n_{Al}=x(mol)$
$n_{Mg}=y(mol)$
$⇒27x+24y=5,1(g)(1)$
$n_{N_2}=\frac{1,12}{22,4}=0,05(mol)$
$Al^0→Al^{+3}+3e$
x → 3x (mol)
$Mg^0→Mg^{+2}+2e$
y → 2y (mol)
$2N^{+5}+10e→N^0_2$
0,5 ← 0,05 (mol)
-Bảo toàn e: ⇒$3x+2y=0,5(2)$
-Từ $(1)$ và $(2)$,ta có hệ pt:
$\left \{ {{27x+24y=5,1} \atop {3x+2y=0,5}} \right.$ $\left \{ {{x=0,1} \atop {y=0,1}} \right.$
-Bảo toàn nguyên tố:
$n_{Al(NO_3)_3}=n_{Al}=0,1(mol)$
$n_{Mg(NO_3)_2}=n_{Mg}=0,1(mol)$
$⇒m_{muối}=m_{Al(NO_3)_3}+m_{Mg(NO_3)_2}$
$=0,1.213+0,1.148=36,1(g)$