Đáp án:
\( {V_{C{O_2}}} = 6,72{\text{ lít}}\)
\({C_{M{\text{ C}}{{\text{H}}_3}COOH}} = 3,2M\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + {K_2}C{O_3}\xrightarrow{{}}2C{H_3}COOK + C{O_2} + {H_2}O\)
Ta có:
\({n_{{K_2}C{O_3}}} = \frac{{41,4}}{{39.2 + 12 + 16.3}} = 0,3{\text{ mol = }}{{\text{n}}_{C{O_2}}}\)
\( \to {V_{C{O_2}}} = 0,3.22,4 = 6,72{\text{ lít}}\)
Trung hòa axit dư
\(2C{H_3}COOH + Ba{(OH)_2}\xrightarrow{{}}{(C{H_3}COO)_2}Ba + 2{H_2}O\)
\( \to {n_{Ba{{(OH)}_2}}} = 0,05.2 = 0,1{\text{ mol}}\)
\( \to {n_{C{H_3}COOH}} = 2{n_{{K_2}C{O_3}}} + 2{n_{Ba{{(OH)}_2}}} = 0,3.2 + 0,1.2 = 0,8{\text{ mol}}\)
\( \to {C_{M{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{0,8}}{{0,25}} = 3,2M\)