Hòa tan hết 11,68 gam hỗn hợp gồm kim loại M và oxit MxOy, trong dung dịch

Đáp án D

 $$\begin{gathered} 11,68\left\{\begin{array}{l}M:a\\ O:b\end{array}\right.\to aM+16b=11,68\left(1\right) \\ M\to M^{n+}+ne \stackrel{+6}{S}+4e\to \stackrel{+4}{S} \\ O+2e\to O^{-2} \end{gathered}$$  

® BT e: $an=0,08.2+2b\left(2\right)$

$$\begin{gathered} n_{H_{2}S{O}_4\text{ phản ứng}}=0,3.\frac{100}{125}=0,24mol \\ \to n_{H_{2}S{O}_4\text{ dư}}=0,06 \\ X\left\{\begin{array}{l}{M}^{n+}\\ S{O}_4^{2-}:n_{S{O}_4^{2-}}=n_{H_{2}S{O}_4}-n_{S{O}_2}=0,16\left(BT\text{ S}\right)\end{array}\right. \end{gathered}$$

® BTĐT: $an=0,16.2\left(3\right)$

$$\begin{gathered} \to \left\{\begin{array}{l}an=0,32\\ b=0,08\\ aM=10,4\end{array}\right.\to \frac{M}{n}=\frac{10,4}{0,32}=32,5\to M=32,5n\to \left\{\begin{array}{l}n=2\\ M=65\left(Zn\right)\end{array}\right. \\ \to n_{ZnS{O}_4}=n_{Zn}=a=0,16 \\ Z{n}^{2+}+4O{H}^{-}\to Zn{O}_2^{2-}+2H_{2}O \\ {H}^{+}+O{H}^{-}\to H_{2}O \\ {n}_{NaOH}=4n_{Zn}+2n_{H_{2}S{O}_4\text{ dư}}=0,76\to V_{NaOH}=760\left(ml\right) \end{gathered}$$.