Hỗn hợp X gồm: Na, Ca,Na2O và CaO. Hoàn tan hết 5,13 gam hỗn hợp X

* Hướng dẫn giải

Đáp án A

$n_{NaOH}=\frac{2,8}{40}=0,07 \mathrm{mol};{\mathrm{n}}_{H_2}=\frac{0,56}{22,4}=0,025\mathrm{mol};{\mathrm{n}}_{S{O}_2}=\frac{1,792}{22,4}=0,08 \mathrm{mol}$

$\left\{\begin{array}{l}\begin{array}{c}\begin{array}{c}Na\stackrel{BTNT.Na}{\to }{n}_{Na}=0,07 \mathrm{mol}\\ Ca:x \mathrm{mol}\end{array}\end{array}\\ O:y \mathrm{mol}\end{array}\right.\to \left\{\begin{array}{l}0,07.23 + 40x + 16y = 5,13\\ \stackrel{BTE}{\to }0,07 + 2x = 2y + 2.0,025\end{array}\right.$

$\to \left\{\begin{array}{l}x=0,06\\ y=0,07\end{array}\right.\to n_{O{H}^{-}\left(Y\right)}=n_{NaOH}+2n_{Ca{\left(OH\right)}_2}=0,07+2.0,06=0,19 \mathrm{mol}$

$\to \frac{n_{O{H}^{-}\left(Y\right)}}{n_{S{O}_2}}=\frac{0,19}{0,08}=2,375 >2\to n_{S{O}_3^{2-}}=n_{S{O}_2}=0,08>{\mathrm{n}}_{C{a}^{2+}}$

$\to CaS{O}_3 tính theo {\mathrm{Ca}}^{2+}\to m=120.0,06=7,2 \mathrm{gam}$