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: 

\({m_{{{(C{H_3}COO)}_2}Mg}} = 0,71{\text{ gam}} \to {{\text{m}}_{{{(C{H_3}COO)}_2}Mg}} = \frac{{0,71}}{{59.2 + 24}} = 0,005{\text{ mol}} \to {{\text{n}}_{C{H_3}COOH}} = 2{n_{{{(C{H_3}COO)}_2}Mg}} = 0,01{\text{ mol}}\)

\( \to {C_{M{\text{ C}}{{\text{H}}_3}COOH}} = \frac{{0,01}}{{0,025}} = 0,4M\)

\({n_{{H_2}}} = {n_{{{(C{H_3}COO)}_2}Mg}} = 0,005{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,005.22,4 = 0,0112{\text{ lít}}\)

\(C{H_3}COOH + NaOH\xrightarrow{{}}C{H_3}COONa + {H_2}O\)

\( \to {n_{NaOH}} = {n_{C{H_3}COOH}} = 0,01{\text{ mol}} \to {{\text{V}}_{NaOH}} = \frac{{0,01}}{{0,05}} = 0,2{\text{ lít}}\)

Đáp án:

Giải thích các bước giải:

a.

$n_{(CH3COO)_2Mg} = \frac{0,71}{142} = 0,005(mol)$

$2CH_3COOH + Mg → (CH_3COO)_2Mg + H_2$

Theo PT trên ,

$n_{CH3COOH} = 2n_{(CH3COO)_2Mg} = 0,005.2 = 0,01(mol)$

$⇒ C_M CH_3COOH = \frac{0,01}{0,025} = 0,4M$

b.

$n_{H2} = n_{(CH3COO)_2Mg} = 0,005(mol)$
$⇒ V_{H2} = 0,005.22,4 = 0,112(lít)$

c.

$CH_3COOH + NaOH → CH_3COONa + H_2O$
$⇒ n_{NaOH} = 0,01(mol)$
$⇒ V_{NaOH} =\frac{0,01}{0,05} = 0,2(lít)$