Đáp án D
$11,68\left\{ \begin{array}{l} M:a\\ O:b \end{array} \right. \to aM + 16b = 11,68{\rm{ }}\left( 1 \right)$
BT e:
$X\left\{ \begin{array}{l} {M^{n + }}\\ SO_4^{2 - }:{n_{SO_4^{2 - }}} = {n_{{H_2}S{O_4}}} - {n_{S{O_2}}} = 0,16\left( {BT{\rm{ S}}} \right) \end{array} \right.$
BTĐT: $an=0,16.2\text{ }\left( 3 \right)$
$\begin{array}{l} \to \left\{ \begin{array}{l} an = 0,32\\ b = 0,08\\ aM = 10,4 \end{array} \right.\\ \to \frac{M}{n} = \frac{{10,4}}{{0,32}} = 32,5\\ \to M = 32,5n\\ \to \left\{ \begin{array}{l} n = 2\\ M = 65\left( {Zn} \right) \end{array} \right. \end{array}$
$\to {{n}_{ZnS{{O}_{4}}}}={{n}_{Zn}}=a=0,16$
$Z{{n}^{2+}}\text{ }+\text{ }4O{{H}^{-}}\text{ }\to \text{ }ZnO_{2}^{2-}\text{ }+\text{ }2{{H}_{2}}O$
${{H}^{+}}\text{ }+\text{ }O{{H}^{-}}\text{ }\to \text{ }{{H}_{2}}O$
${{n}_{NaOH}}=4{{n}_{Zn}}+2{{n}_{{{H}_{2}}S{{O}_{4}}\text{ d }\!\!\ddot{\mathrm{o}}\!\!\text{ }}}=0,76\to {{V}_{NaOH}}=760\left( ml \right)$.
