Đáp án:

Giải thích các bước giải:Gọi n_Na2CO3 = x (mol)

 $Na_{2}$ $CO_{3}$ + 2HCl →→ 2NaCl + $H_{2}$ O + $Co_{2}$ 

x →→ 2x →→ 2 x (mol)

C%(NaCl) =$\frac{2,58,5x}{200+120}$ . 100% = 20%

=> x =0,547 (mol)

m_Na2CO3 = 0,547 . 106 = 57,982 (g)

m_HCl = 2 . 0,547 . 36,5 =39,931 (g)

C%_(Na2CO3) =$\frac{57,892}{200}$ . 100% = 28,946%

C%(HCl) = $\frac{39,931}{120}$ . 100% = 33,28%

Em tham khảo nha :

\(\begin{array}{l}
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
{n_{N{a_2}C{O_3}}} = \dfrac{{120 \times 21,2\% }}{{106}} = 0,24mol\\
{n_{HCl}} = \dfrac{{200 \times 14,6\% }}{{36,5}} = 0,8mol\\
\dfrac{{0,24}}{1} < \dfrac{{0,8}}{2} \Rightarrow HCl\text{ dư}\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,24mol\\
{V_{C{O_2}}} = 0,24 \times 22,4 = 5,376l\\
b)\\
{n_{NaCl}} = 2{n_{N{a_2}C{O_3}}} = 0,48mol\\
{n_{HC{l_d}}} = 0,8 - 0,24 \times 2 = 0,32mol\\
C{\% _{NaCl}} = \dfrac{{0,48 \times 58,5}}{{120 + 200 - 0,24 \times 44}} \times 100\%  = 9,1\% \\
C{\% _{HC{l_d}}} = \dfrac{{0,32 \times 36,5}}{{120 + 200 - 0,24 \times 44}} \times 100\%  = 3,8\% 
\end{array}\)