Đáp án:
\({{\text{m}}_{dd{\text{ C}}{{\text{H}}_3}COOH}} = 200{\text{ gam}}\)
\({V_{C{O_2}}} = 4,48{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + CaC{O_3}\xrightarrow{{}}{(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O\)
Ta có:
\({n_{CaC{O_3}}} = \frac{{20}}{{100}} = 0,2{\text{ mol = }}{{\text{n}}_{C{O_2}}} \to {n_{C{H_3}COOH}} = 2{n_{CaC{O_3}}} = 0,2.2 = 0,4{\text{ mol}}\)
\( \to {m_{C{H_3}COOH}} = 0,4.60 = 24{\text{ gam}} \to {{\text{m}}_{dd{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{24}}{{12\% }} = 200{\text{ gam}}\)
\({V_{C{O_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)