Ans:

+ Số mol ban đầu là:

☆ $n_{\text{Ba}} = \displaystyle\frac{27{,}4}{137} = 0{,}2 \text{ mol}$

☆ $m_{\text{H}_2\text{SO}_4} = \displaystyle\frac{100 \times 9{,}8\%}{100\%} = 9{,}8 \text{ g} \Rightarrow n_{\text{H}_2\text{SO}_4} = \displaystyle\frac{9{,}8}{98} = 0{,}1 \text{ mol}$

+ Phương trình phản ứng là:

☆ $\text{Ba} + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + \text{H}_2 \uparrow$

☆ $0{,}1 \rightarrow 0{,}1 \rightarrow 0{,}1 \rightarrow 0{,}1 \text{ (mol)}$

☆ $\text{Ba dư} + 2\text{H}_2\text{O} \rightarrow \text{Ba(OH)}_2 + \text{H}_2 \uparrow$

☆ $0{,}1 \rightarrow 0{,}1 \rightarrow 0{,}1 \text{ (mol)}$

+ Thể tích khí $\text{H}_2$ là:

☆ $n_{\text{H}_2} = 0{,}1 + 0{,}1 = 0{,}2 \text{ mol}$

☆ $V_{\text{H}_2} = 0{,}2 \times 24{,}79 = 4{,}958 \text{ lít}$

+ Nồng độ phần trăm dung dịch sau phản ứng là:

☆ $m_{\text{Ba(OH)}_2} = 0{,}1 \times 171 = 17{,}1 \text{ g}$

☆ $m_{\text{BaSO}_4} = 0{,}1 \times 233 = 23{,}3 \text{ g}$

☆ $m_{\text{H}_2} = 0{,}2 \times 2 = 0{,}4 \text{ g}$

☆ $m_{\text{dd sau pư}} = 27{,}4 + 100 - 23{,}3 - 0{,}4 = 103{,}7 \text{ g}$

☆ $C\%_{\text{Ba(OH)}_2} = \displaystyle\frac{17{,}1}{103{,}7} \times 100\% \approx 16{,}49\%$

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