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%