Nung hỗn hợp X gồm Al, Fe và Cu (trong đó Cu chiếm 34,72% khối lượng) trong không khí một thời gian, thu được 6,17 gam hỗn hợp rắn Y gồm các kim loại và oxit tương ứng. Cho hỗn hợp...

$X\left\{\begin{array}{l}Al\\ Fe\\ Cu\end{array}\right.\underset{KK}{\overset{t°}{\to }}\underset{6,17\left(g\right)}{\underbrace{\text{raén Y}\left.〈\begin{array}{l}Oxit\text{ KL}\\ \text{KL dö}\end{array}\right|}}+A\left\{\begin{array}{l}KHS{O}_4:0,36\left(mol\right)\\ KN{O}_3:0,04\left(mol\right)\end{array}\right.〈\begin{array}{l}\underset{56,05\left(g\right)}{\underbrace{B\left\{\begin{array}{l}C{u}^{2+}\\ F{e}^{3+}\\ A{l}^{3+}\\ N{H}_4^{+}:0,02\left(mol\right)\\ K^{+}:0,4\left(mol\right)\\ S{O}_4^{2-}\end{array}\right.}}\\ \underset{0,6\left(gam\right)}{\underbrace{\text{Khí Z}\left(Oxit\text{ Nitô}\right)}}:0,015\left(mol\right)\end{array}+\underset{2,52\left(gam\right)}{\underbrace{H_{2}O}}$· $BTKL:m_Y+m_A=m_B+m_Z+m_{H_{2}O}\to m_{H_{2}O}=2,52\left(gam\right)\to n_{H_{2}O}=0,14\left(mol\right)$

· $BTH:n_{KHS{O}_4}=4n_{N{H}_4^{+}}+2n_{H_{2}O}\to n_{N{H}_4^{+}}=\frac{0,36-2.0,14}{4}=0,02\left(mol\right)$

· $$\begin{gathered} BTN:n_{N\left(Z\right)}=n_{KN{O}_3}-n_{N{H}_4^{+}}=0,02\left(mol\right);m_Z=m_{N\left(Z\right)}+m_{O\left(Z\right)}=0,6\left(gam\right) \\ \to m_{O\left(Z\right)}=0,32\left(gam\right)\to n_{O\left(Z\right)}=0,02\left(mol\right) \end{gathered}$$

· $BTO:n_{O\left(Y\right)}+3n_{KN{O}_3}=n_{O\left(Z\right)}+n_{H_{2}O}\to n_{O\left(Y\right)}=0,04\left(mol\right)$

·  $m_{KL\left(X\right)}=m_Y-m_{O\left(Y\right)}=6,17-0,04.16=5,53\left(gam\right)$                      (1)

Do Cu chiếm 34,72% về khối lượng trong X $\to m_{Cu}=1,92\left(gam\right)\to n_{Cu}=0,03\left(mol\right)$

· $BTe:n_{e\left(KL\right)}=2n_{S{O}_4^{2-}}-n_{K^{+}}-n_{N{H}_4^{+}}=2.0,36-0,4-0,02=0,3\left(mol\right)$     (2)

Từ (1), (2): $\left\{\begin{array}{l}56a+27b+\underset{m_{Ca}}{\underbrace{1,92}}=5,53\\ 3a+3b+\underset{n_{e\left(Cu\right)}}{\underbrace{0,03.2}}=0,3\end{array}\right.\to \left\{\begin{array}{l}a=0,05=n_{Fe}\\ b=0,03=n_{Al}\end{array}\right.$

· $B+NaOH$  ® lượng $O{H}^{-}$  dùng để kết hợp các cation  là: $\left(C{u}^{2+},F{e}^{3+},A{l}^{3+},N{H}_4^{+}\right)$

$n_{O{H}^{-}}=\underset{n_{A{l}^{3+}}}{\underbrace{0,03.3}}+\underset{n_{F{e}^{3+}}}{\underbrace{0,05.3}}+\underset{n_{C{u}^{2+}}}{\underbrace{0,03.2}}+\underset{n_{N{H}_4^{+}}}{\underbrace{0,02}}=0,32\left(mol\right)\to n_{O{H}^{-}\text{dö}}=\underset{\text{ban ñaàu}}{\underbrace{0,34}}-\underset{\left(\text{pö}\right)}{\underbrace{0,32}}=0,02\left(mol\right)$

·  $0,02molO{H}_{\text{dö}}^{-}$ hòa tan $Al{\left(OH\right)}_3\to Al{\left(OH\right)}_4^{-}\to n_{Al{\left(OH\right)}_3\text{ dö}}=0,03-0,02=0,01\left(mol\right)$

$\Rightarrow m_{\text{keát tuûa}}=\underset{0,05.107}{\underbrace{m_{Fe{\left(OH\right)}_3}}}+\underset{0,03.98}{\underbrace{m_{Cu{\left(OH\right)}_2}}}+\underset{0,01.78}{\underbrace{m_{Al{\left(OH\right)}_{3\left(\text{dö}\right)}}}}=9,07\left(gam\right)$

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Chọn đáp án D.