Volume of NaOH required to react = 145.5 ml
Reaction
CO₂(g) + 2 NaOH(aq) ⇒Na₂CO₃(aq) + H₂O(l)
The volume of CO₂ : 0.45 L
mol CO₂ at STP (O C, 1 atm) ⇒ at STP 1 mol gas 22.4 L :
[tex]\tt mol~CO_2=\dfrac{0.45}{22.4}=0.02[/tex]
From the equation, the mol ratio of CO₂ : NaOH = 1 : 2, so mol NaOH :
[tex]\tt mol~NaOH=\dfrac{2}{1}\times 0.02=0.04[/tex]
Then volume of NaOH :
[tex]\tt V=\dfrac{n}{M}\\\\V=\dfrac{0.04}{0.275}\\\\V=0.1455~L\rightarrow 145.5~mL[/tex]
