Respuesta :
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
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 : The mass of [tex]Na_2CO_3[/tex] prepared can be 3.18 grams.
Explanation : Given,
Mass of NaOH = 2.40 g
Molar mass of NaOH = 40 g/mole
Molar mass of [tex]Na_2CO_3[/tex] = 106 g/mole
First we have to calculate the moles of [tex]NaOH[/tex].
[tex]\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{2.40g}{40g/mole}=0.06moles[/tex]
Now we have to calculate the moles of [tex]Na_2CO_3[/tex].
The balanced chemical reaction is,
[tex]2NaOH(s)+CO_2(g)\rightarrow Na_2CO_3(s)+H_2O(l)[/tex]
From the balanced reaction we conclude that,
As, 2 moles of [tex]NaOH[/tex] react to give 1 mole of [tex]Na_2CO_3[/tex]
So, 0.06 moles of [tex]NaOH[/tex] react to give [tex]\frac{0.06}{2}=0.03moles[/tex] of [tex]Na_2CO_3[/tex]
Now we have to calculate the mass of [tex]Na_2CO_3[/tex].
[tex]\text{Mass of }Na_2CO_3=\text{Moles of }Na_2CO_3\times \text{Molar mass of }Na_2CO_3[/tex]
[tex]\text{Mass of }Na_2CO_3=(0.03mole)\times (106g/mole)=3.18g[/tex]
Therefore, the mass of [tex]Na_2CO_3[/tex] prepared can be 3.18 grams.
