Đáp án:

\({m_{BaC{{\text{O}}_3}}} = 14,775{\text{ gam}}\)

Giải thích các bước giải:

Ta có:

\({n_{KOH}} = 0,1.1 = 0,1{\text{ mol;}}{{\text{n}}_{Ba{{(OH)}_2}}} = 0,1.075 = 0,075{\text{ mol}}\)

\( \to {n_{O{H^ - }}} = {n_{KOH}} + 2{n_{Ba{{(OH)}_2}}} = 0,1 + 0,075.2 = 0,25{\text{ mol}}\)

\({n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1{\text{ mol}} \\\to \dfrac{{{n_{O{H^ - }}}}}{{n{ _{C{O_2}}}}} = \dfrac{{0,25}}{{0,1}} = 2,5 > 2\)

Phản ứng:

\(2O{H^ - } + C{O_2}\xrightarrow{{}}C{O_3}^{2 - } + {H_2}O\)

\(B{a^{2 + }} + C{O_3}^{2 - }\xrightarrow{{}}BaC{{\text{O}}_3}\)

\({n_{C{O_3}^{2 - }}} = {n_{C{O_2}}} = 0,1{\text{ mol > }}{{\text{n}}_{B{a^{2 + }}}}\)

\( \to {n_{BaC{{\text{O}}_3}}} = {n_{B{a^{2 + }}}} = 0,075{\text{ mol}}\)

\( \to {m_{BaC{{\text{O}}_3}}} = 0,075.197 = 14,775{\text{ gam}}\)