2NaOH + CO2 \(\rightarrow\)Na2CO3 + H2O
mdd NaOH=\(156.1,22=190,32\left(g\right)\)
mNaOH=\(190,32.\dfrac{20}{100}=38,064\left(g\right)\)
nCO2=\(\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PTHH ta có:
nCO2=nNa2CO3=0,25(mol)
2nCO2=nNaOH=0,5(mol)
mNa2CO3=0,25.106=26,5(g)
mNaOH(tác dụng)=0,5.40=20(g)
mNaOH còn lại=38,064-20=18,064(g)