Đáp án:
a) \({{\text{C}}_{M\;{\text{C}}{{\text{H}}_3}COOH}} = 0,3M\)
b) \({{\text{V}}_{C{O_2}}} = 1,68{\text{ lít}}\)
Giải thích các bước giải:
a) Phản ứng xảy ra:
\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)
Ta có:
\({m_{NaOH}} = 30.20\% = 6{\text{ gam}} \to {{\text{n}}_{NaOH}} = {n_{C{H_3}COOH}} = \frac{6}{{40}} = 0,15{\text{ mol}} \to {{\text{C}}_{M\;{\text{C}}{{\text{H}}_3}COOH}} = \frac{{0,15}}{{0,5}} = 0,3M\)
b) \({n_{C{H_3}COOH}} = 0,15{\text{ mol; }}{{\text{n}}_{N{a_2}C{O_3}}} = 0,2.0,5 = 0,1{\text{ mol}}\)
Phản ứng xảy ra:
\(2C{H_3}COOH + N{a_2}C{O_3}\xrightarrow{{}}2C{H_3}COONa + C{O_2} + {H_2}O\)
\({n_{C{H_3}COOH}} < 2{n_{N{a_2}C{O_3}}} \to N{a_2}C{O_3}\) dư.
\( \to {n_{C{O_2}}} = \frac{1}{2}{n_{C{H_3}COOH}} = 0,075{\text{ mol}} \to {{\text{V}}_{C{O_2}}} = 0,075.22,4 = 1,68{\text{ lít}}\)
