Đáp án:
V=4,48 lít; m=23,3 gam
Giải thích các bước giải:
\({n_{Ba}} = \frac{{27,4}}{{137}} = 0,2{\text{ mol;}}{{\text{m}}_{{H_2}S{O_4}}} = 400.2,45\% = 9,8{\text{ gam}} \to {{\text{n}}_{{H_2}S{O_4}}} = \frac{{9,8}}{{98}} = 0,1{\text{mol < }}{{\text{n}}_{Ba}}\)
Phản ứng xảy ra:
\(Ba + {H_2}S{O_4}\xrightarrow{{}}BaS{O_4} + {H_2}\)
\(Ba + 2{H_2}O\xrightarrow{{}}Ba{(OH)_2} + {H_2}\)
Ta có:
\({n_{{H_2}}} = {n_{Ba}} = 0,2{\text{ mol}} \to {\text{V = 0}}{\text{,2}}{\text{.22}}{\text{,4 = 4}}{\text{,48 lít}}\)
\({n_{BaS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,1{\text{ mol}} \to {\text{m = 0}}{\text{,1}}{\text{.233 = 23}}{\text{,3 gam}}\)
BTKL: \({m_{dd{\text{sau phản ứng}}}} = 27,4 + 400 - 0,2.2 = 427{\text{ gam}}\)
\({n_{Ba{{(OH)}_2}}} = {n_{Ba}} - {n_{{H_2}S{O_4}}} = 0,2 - 0,1 = 0,1{\text{ mol}} \to {{\text{m}}_{Ba{{(OH)}_2}}} = 0,1.(137 + 17.2) = 17,1{\text{ gam}} \to {\text{\% }}{{\text{m}}_{Ba{{(OH)}_2}}} = \frac{{17,1}}{{427}} = 4\% \)
