Question

Answer 1

Ta có: 27n_Al + 56n_Fe + 64n_Cu = 23,82 (1)

\(n_{H_2}=\dfrac{9,408}{22,4}=0,42\left(mol\right)\)

BT e, có: 3n_Al + 2n_Fe = 2n_H2 = 0,42.2 (2)

Có: n_HNO3 = 0,8.2 = 1,6 (mol)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2}+n_{NO}=\dfrac{3,584}{22,4}=0,16\\12n_{N_2}+4n_{NO}=n_{HNO_3}=1,6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{N_2}=0,12\left(mol\right)\\n_{NO}=0,04\left(mol\right)\end{matrix}\right.\)

BT e, có: 3n_Al + 3n_Fe + 2n_Cu = 10n_N2 + 3n_NO = 1,32 (3)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,18\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\\n_{Cu}=0,165\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{Al}=\dfrac{0,18.27}{23,82}.100\%\approx20,4\%\)

→ Đáp án: B