Question

How many particle or molecules of `Na_2CO_3` are produced when 10g of `NaOH` is used? `22.3 x 10^24` particles of `NaOH` are needed to produce how many liters of `H_2O`? How many particles/molecules of `NaOH` are from 30 gram of `NaOH`?

`2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)`

Answer 1

a) `7.5 xx 10^22` molecules of `Na_2CO_3` are produced
b)cannot be answered as stoichiometry deals with gases not liquids and water is liquid
c)There are `4.5xx 10^23` molecules of `NaOH` in `30 g`

Explanation:

`2NaOH(s) + CO_2(g) -> Na_2CO_3(s) + H_2O(l)`

Molar mass of `NaOH = 40`

`2xx40 g` of `NaOH` produces `6.02 xx 10^23` molecules of `Na_2CO_3`
`:.` `10 g` of `NaOH` produces `(6.02xx10^23xx10)/(2xx40)` molecules of `Na_2CO_3`
`= 7.5 xx 10^22` molecules of `Na_2CO_3`

`40g` of `NaOH` contains `6xx10^23` molecules
`:.` `30g` of `NaOH` contains `(30xx6xx10^23)/40` molecules
`=4.5xx10^23` molecules