Question

Sodium hydroxide reacts with carbon dioxide to form sodium carbonate and water: 2 naoh(s) + co2(g) → na2co3(s) + h2o(l) how many grams of na2co3 can be prepared from 2.40 g of naoh?

Answer 1

no of moles of naoh = 2.40 ÷ (23+16+1) = 0.06mol

no of moles of na2co3 = 0.06 ÷ 2 = 0.03mol

mass of na2co3 = 0.03 × (23×2+12+16×3) = 0.03 × 106 = 3.18g

Answer 2

Answer : The mass of
`Na_2CO_3` prepared can be 3.18 grams.

Explanation : Given,

Mass of NaOH = 2.40 g

Molar mass of NaOH = 40 g/mole

Molar mass of
`Na_2CO_3` = 106 g/mole

First we have to calculate the moles of
`NaOH`.

`\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=(2.40g)/(40g/mole)=0.06moles`

Now we have to calculate the moles of
`Na_2CO_3`.

The balanced chemical reaction is,

`2NaOH(s)+CO_2(g)\rightarrow Na_2CO_3(s)+H_2O(l)`

From the balanced reaction we conclude that,

As, 2 moles of
`NaOH` react to give 1 mole of
`Na_2CO_3`

So, 0.06 moles of
`NaOH` react to give
`(0.06)/(2)=0.03moles` of
`Na_2CO_3`

Now we have to calculate the mass of
`Na_2CO_3`.

`\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3* \text{Molar mass of }Na_2CO_3`

`\text{Mass of }Na_2CO_3=(0.03mole)* (106g/mole)=3.18g`

Therefore, the mass of
`Na_2CO_3` prepared can be 3.18 grams.