Đáp án:
\({V_{KMn{O_4}}} = 0,576{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(5{C_6}{H_4}{(C{H_3})_2} + 12KMn{O_4} + 18{H_2}S{O_4}\xrightarrow{{{t^o}}}5{C_6}{H_4}{(COOH)_2} + 6{K_2}S{O_4} + 12MnS{O_4} + 28{H_2}O\)
Ta có:
\({n_{{C_6}{H_4}{{(C{H_3})}_2}}} = \frac{{10,6}}{{12.8 + 10}} = 0,1{\text{ mol}} \to {{\text{n}}_{KMn{O_4}}} = \frac{{12}}{5}{n_{{C_6}{H_4}{{(C{H_3})}_2}}} = 0,24{\text{ mol}} \to {{\text{n}}_{KMn{O_4}{\text{ tham gia}}}} = 0,24.120\% = 0,288{\text{ mol}}\)
\( \to {V_{KMn{O_4}}} = \frac{{0,288}}{{0,5}} = 0,576{\text{ lít}}\)