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 $NaOH$.

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

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 $\frac{0.06}{2}=0.03moles$ of $Na_2CO_3$

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

$\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3$

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

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