Đáp án:
a) \({C_{M{\text{ C}}{{\text{H}}_3}COOH}} = 0,4M\)
b) \({{\text{V}}_{{H_2}}} = 0,0112{\text{ lít}}\)
c) \({{\text{V}}_{NaOH}} = 0,2{\text{ lít}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(2C{H_3}COOH + Mg\xrightarrow{{}}{(C{H_3}COO)_2}Mg + {H_2}\)
Ta có:
\({m_{{{(C{H_3}COO)}_2}Mg}} = 0,71{\text{ gam}} \to {{\text{m}}_{{{(C{H_3}COO)}_2}Mg}} = \frac{{0,71}}{{59.2 + 24}} = 0,005{\text{ mol}} \to {{\text{n}}_{C{H_3}COOH}} = 2{n_{{{(C{H_3}COO)}_2}Mg}} = 0,01{\text{ mol}}\)
\( \to {C_{M{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{0,01}}{{0,025}} = 0,4M\)
\({n_{{H_2}}} = {n_{{{(C{H_3}COO)}_2}Mg}} = 0,005{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,005.22,4 = 0,0112{\text{ lít}}\)
\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)
\( \to {n_{NaOH}} = {n_{C{H_3}COOH}} = 0,01{\text{ mol}} \to {{\text{V}}_{NaOH}} = \frac{{0,01}}{{0,05}} = 0,2{\text{ lít}}\)
