II.Tự luận
Câu 1:
1.$N_2+O_2\xrightarrow{300^oC}2NO ↑$
2.$2NO+O_2→2NO_2↑$
3.$4NO_2+O_2+2H_2O→4HNO_3$
4.$NH_3+HNO_3→NH_4NO_3$
5.$Na_4NO_3+NaOH→NaNO_3+NH_3↑+H_2O$
6.$2NH_3+H_2SO_4→(NH_4)_2SO_4$
Câu 2:
a.
Gọi $n_{Al}=x(mol)$
$n_{Fe}=y(mol)$
$⇒27x+56y=11(g)(1)$
$n_{NO}=\frac{6,72}{22,4}=0,3(mol)$
$Al^0→Al^{+3}+3e$
x → 3x (mol)
$Fe^0→Fe^{+3}+3e$
y → 3y (mol)
$N^{+5}+3e→N^{+2}$
0,9 ← 0,3 (mol)
-Bảo toàn e: ⇒ $3x+3y=0,9(2)$
Từ $(1)$ và $(2)$, ta có hệ pt:
$\left \{ {{27x+56y=11} \atop {3x+3y=0,9}} \right.$ $\left \{ {{x=0,2} \atop {y=0,1}} \right.$
-%$m_{Fe}=\frac{0,1.56}{11}.100$%$≈50,9$%
-%$m_{Al}=100$%$-$%$m_{Fe}=100$%$-50,9$%$=49,1$%
b.
-Bảo toàn $Al$ :⇒ $n_{Al}=n_{Al(NO_3)3}=0,2(mol)$
-Bảo toàn $Fe$ :⇒ $n_{Fe}=n_{Fe(NO_3)3}=0,1(mol)$
$⇒m_{muối}=m_{Al(NO_3)_3}+m_{Fe(NO_3)_3}$
$=0,2.213+0,1.242=66,8(g)$
c.
-$n_{HNO_3}=3.n_{Al(NO_3)_3}+3.n_{Fe(NO_3)_3}+n_{NO}$
$=3.0,2+3.0,1+0,3=1,2(mol)$
$⇒V_{dd..HNO_3}=\frac{1,2}{2}=0,6(l)$
