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.

\( \to {n_{C{O_2}}} = \frac{1}{2}{n_{C{H_3}COOH}} = 0,075{\text{ mol}} \to {{\text{V}}_{C{O_2}}} = 0,075.22,4 = 1,68{\text{ lít}}\)

Đáp án:

Giải thích các bước giải:

 a)CH3COOH+NaOH-->CH3COONa+H2O

n NaOH=30.20%/40=0,15(mol)

n CH3COOH=n NaOH=0,15(mol)

CM CH3COOH=0,15/0,5=0,3(M)

b)2CH3COOH+Na2CO3-->2CH3COONa+H2O+CO2

n na2CO3=0,2.0,5=0,1(mol)

=>Na2CO3 dư

n CO2=1/2 n CH3COOH=0,075(mol)

V CO2=0,075.22,4=1,68(l)