a,
Gọi a, b là mol Al, Fe.
$\Rightarrow 27a+56b=8,3$ (1)
Bảo toàn e: $3a+3b=0,1.3+0,03.10=0,6$ (2)
(1)(2) $\Rightarrow a=b=0,1$
$\%m_{Al}=\frac{0,1.27.100}{8,3}=32,53\%$
$\%m_{Fe}=67,47\%$
b,
Bảo toàn nguyên tố:
$n_{Al(NO_3)_3}= n_{Al}= a$
$n_{Fe(NO_3)_3}=n_{Fe}= b$
$\Rightarrow m_{\text{muối}}= 213.0,1+242.0,1=45,5g$
c,
Bảo toàn N:
$n_{HNO_3}= 3n_{Al(NO_3)_3}+ 3n_{Fe(NO_3)_3}+ n_{NO}+ 2n_{N_2}$
$= 0,76 mol$