Đáp án:
\({\text{C}}{{\text{\% }}_{Ba{{(OH)}_2}}} = 16,3\% \)
Giải thích các bước giải:
Ta có:
\({n_{Ba}} = \frac{{68,5}}{{137}} = 0,5{\text{ mol;}}{{\text{m}}_{{H_2}S{o_4}}} = 200.14,7 = 29,4{\text{ gam}} \to {{\text{n}}_{{H_2}S{O_4}}} = \frac{{29,4}}{{98}} = 0,3{\text{ mol < }}{{\text{n}}_{Ba}}\)
Phản ứng xảy ra:
\(Ba + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} \downarrow + {H_2}\)
\(Ba + 2{H_2}O\xrightarrow{{}}Ba{(OH)_2} + {H_2}\)
Ta có:
\({n_{BaS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,3{\text{ mol;}}{{\text{n}}_{Ba{{(OH)}_2}}} = 0,5 - 0,3 = 0,2{\text{ mol;}}{{\text{n}}_{{H_2}}} = {n_{Ba}} = 0,5{\text{ mol}}\)
BTKL:
\({m_{dd{\text{sau phản ứng}}}} = {m_{Ba}} + {m_{dd{\text{ }}{{\text{H}}_2}S{O_4}}} - {m_{BaS{O_4}}} - {m_{{H_2}}} = 68,5 + 200 - 0,3.233 - 0,5.2 = 197,6{\text{ gam}}\)
\({m_{Ba{{(OH)}_2}}} = 0,2{\text{ mol}} \to {{\text{m}}_{Ba{{(OH)}_2}}} = 0,2.(137 + 17.2) = 32,2{\text{ gam}} \to {\text{C}}{{\text{\% }}_{Ba{{(OH)}_2}}} = \frac{{32,2}}{{197,6}} = 16,3\% \)
