Question #a6aa0

1 Answer
P dilip_k
Jan 30, 2017

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.

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