Em tham khảo nah :

\(\begin{array}{l}
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
{m_{N{a_2}C{O_3}}} = \dfrac{{200 \times 21,2}}{{100}} = 42,4g\\
{n_{N{a_2}C{O_3}}} = \dfrac{{42,4}}{{106}} = 0,4mol\\
{n_{HCl}} = \dfrac{{120 \times 14,6}}{{100}} = 17,52g\\
{n_{HCl}} = \dfrac{{17,52}}{{36,5}} = 0,48mol\\
\dfrac{{0,4}}{1} > \dfrac{{0,48}}{2} \Rightarrow N{a_2}C{O_3}\text{ dư}\\
{n_{C{O_2}}} = \dfrac{{{n_{HCl}}}}{2} = 0,24mol\\
{V_{C{O_2}}} = 0,24 \times 22,4 = 5,376g\\
b)\\
{n_{NaCl}} = {n_{HCl}} = 0,48mol\\
{m_{NaCl}} = 0,48 \times 58,5 = 28,08g\\
{n_{N{a_2}C{O_3}d}} = {n_{N{a_2}C{O_3}}} - \dfrac{{{n_{HCl}}}}{2} = 0,16mol\\
{m_{N{a_2}C{O_3}}} = 0,16 \times 106 = 16,96g\\
{m_{{\rm{dd}}spu}} = 200 + 120 - 0,24 \times 44 = 309,44g\\
C{\% _{NaCl}} = \dfrac{{28,08}}{{309,44}} \times 100\%  = 9,07\% \\
C{\% _{N{a_2}C{O_3}}} = \dfrac{{16,96}}{{309,44}} \times 100\%  = 5,5\% 
\end{array}\)

Na2CO3+2HCl→2NaCl+H2O+CO2

mNa2CO3=200*21,2/100=42,4g

nNa2CO3=42,4/106=0,4mol

mHCl=120*14,6/100=17,52g

nHCl=17,52/36,5=0,48mol

Theo pt: nHCl/nNa2CO3=2/1

Theo đề nHCl/nNa2CO3=1,2/1

Vậy Na2CO3 dư

nCO2=0,48/2=0,24mol

0,24=$\frac{PV}{RT}$ =$\frac{2V}{0,082.(10+273)}$ 

⇔2V=0,082*283*0,24=5,57

⇔V=2,785l

b)mdd sau phản ứng=200+120-0,24*44=309,44g

mNaCl=0,48*58,5=28,08g

C%NaCl=28,08/309,44*100%=9,07%

nNa2CO3 dư=0,4-0,48/2=0,16mol

mNa2CO3 dư=0,16*106=16,96g

C%Na2CO3 dư=16,96/309,44*100%=5,48%