Question #a6aa0

Jan 30, 2017

Reaction of $C {O}_{2}$ with $N a O H$ occurs in following two steps. In first step $N {a}_{2} C {O}_{3}$ is produced and in second step it again reacts with excess $C {O}_{2}$ to produce $N a H C {O}_{3}$

$2 N a O H + C {O}_{2} \to N {a}_{2} C {O}_{3} + {H}_{2} O$

$N {a}_{2} C {O}_{3} + C {O}_{2} + {H}_{2} O \to 2 N a H C {O}_{3}$

As per 1st step reaction we see

1 mol $C {O}_{2}$ reacts with 2 moles or $2 m o l \times 40 g \text{/mol"=80g" } N a O H$

So $0.18 m o l$ will react with $80 \times 0.18 g = 14.4 g \text{ } N a O H$

So NaOH will be excess in the reaction mixture as 16.5g $N a O H$ was taken and in this strong alkaline environment $N {a}_{2} C {O}_{3}$ will be final product.