Bạn tham khảo:
$n_{Fe(NO_3)_2}=0,15(mol)$
$n_{Cu(NO_3)_2}=0,05(mol)$
$a/$
$2N+nFe(NO_3)_2 \to 2N(NO_3)_n+nFe(1)$
$2N+nCu(NO_3)_2 \to 2N(NO_3)_n+nCu(2)$
$(1)$
$n_{Fe}=n_{Fe(NO_3)_2}=0,15(mol)$
$m_{Fe}=0,15.56=8,4(g)$
$(2)$
$n_{Cu}=n_{Cu)NO_3)_2}=0,05(mol)$
$m_{Cu}=0,05.64=3,2(g)$
$3,2 ≤m≤8,4$
$m_N=16,4-8,4-3,2=4,8(g)$
$n_N=\frac{0,15.2}{n}+\frac{0,05.2}{n}=\frac{0,4}{n}(mol)$
$M_N=\frac{4,8.n}{0,4}=12.n$
$n=2; N=24(Mg)$
$b/$
$CM_{Mg(NO_3)_2}=\frac{0,15+0,05}{0,25}=0,8M$