`n_{\text{BaCO}_3} = \frac{19.7}{197} = 0.1 \text{ mol}`

`n_{\text{HNO}_3} = 0.5 \cdot 1 = 0.5 \text{ mol}`

`PTHH:`

`\text{BaCO}_3 + 2\text{HNO}_3 \rightarrow \text{Ba(NO}_3)_2 + \text{CO}_2 \uparrow + \text{H}_2\text{O}`

So sánh tỉ lệ: `0.1 < \frac{0.5}{2} \Rightarrow \text{BaCO}_3` phản ứng hết, `\text{HNO}_3` dư.

Theo `PT`:

`n_{\text{CO}_2} = n_{\text{Ba(NO}_3)_2} = n_{\text{BaCO}_3} = 0.1 \text{ mol}`

`n_{\text{HNO}_3 \text{ pứ}} = 2 \cdot 0.1 = 0.2 \text{ mol}`

`n_{\text{HNO}_3 \text{ dư}} = 0.5 - 0.2 = 0.3 \text{ mol}`

`a)` Thể tích khí `B (\text{CO}_2)`:

`V = 0.1 \cdot 24.79 = 2.479 \text{ lít}`

`b)` Muối trong `A` là: Barium nitrate (`\text{Ba(NO}_3)_2`)

Các chất tan gồm: `\text{Ba(NO}_3)_2` và `\text{HNO}_3` dư.

`C_{M(\text{Ba(NO}_3)_2)} = \frac{0.1}{0.5} = 0.2 \text{M}`

`C_{M(\text{HNO}_3 \text{ dư})} = \frac{0.3}{0.5} = 0.6 \text{M}`