Hòa tan hoàn toàn muối MCO3

A

- Hướng dẫn giải

Đáp án  A

Giả sử $n_{MC{O}_3}$= 1 mol → $n_{H_{2}S{O}_4}=n_{MS{O}_4}=n_{C{O}_2}$= 1mol

→$m_{H_{2}S{O}_4}$= 98 gam; $m_{MS{O}_4}$= M+96 gam;

$m_{C{O}_2}$= 44 gam; $m_{dd H_{2}S{O}_4 12,25\%}$= 800 g

MCO₃+ H₂SO₄→ MSO₄+ CO₂↑ + H₂O

Ta có: m_{dd sau pứ} = $m_{MC{O}_3}+m_{dd H_{2}S{O}_4}-m_{C{O}_2}$

= M+60+800-44 = M+816 (gam)

$$\begin{gathered} C{\%}_{MS{O}_4}=\frac{M+96}{M+816}.100\%=14,29\% \\ \to M=24 \end{gathered}$$

=> M là Mg