Đáp án:
\( {m_{{{(C{H_3}COO)}_2}Ca}} = 30,02{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + Ca{(OH)_2}\xrightarrow{{}}{(C{H_3}COO)_2}Ca + 2{H_2}O\)
Ta có:
\({n_{C{H_3}COOH}} = \frac{{24}}{{60}} = 0,4{\text{ mol}}\)
\( \to {n_{{{(C{H_3}COO)}_2}Ca{\text{ lt}}}} = \frac{1}{2}{n_{C{H_3}COOH}} = 0,2{\text{ mol}}\)
\( \to {n_{{{(C{H_3}COO)}_2}Ca}} = 0,2.95\% = 0,19{\text{ mol}}\)
\( \to {m_{{{(C{H_3}COO)}_2}Ca}} = 0,19.(59.2 + 40) = 30,02{\text{ gam}}\)