Question

Question #a6aa0

Answer 1

Reaction of `CO_2` with `NaOH` occurs in following two steps. In first step `Na_2CO_3` is produced and in second step it again reacts with excess `CO_2` to produce `NaHCO_3`

`2NaOH+CO_2->Na_2CO_3+H_2O`

`Na_2CO_3+CO_2+H_2O->2NaHCO_3`

As per 1st step reaction we see

1 mol `CO_2` reacts with 2 moles or `2molxx40g"/mol"=80g" "NaOH`

So `0.18mol` will react with `80xx0.18g=14.4g" "NaOH`

So NaOH will be excess in the reaction mixture as 16.5g `NaOH` was taken and in this strong alkaline environment `Na_2CO_3` will be final product.