Respuesta :
Answer: The theoretical yield of [tex]Na_2SO_4[/tex] is, 5.65 grams.
Explanation : Given,
Mass of [tex]H_2SO_4[/tex] = 3.9 g
Mass of [tex]NaOH[/tex] = 7.8 g
Molar mass of [tex]H_2SO_4[/tex] = 98 g/mol
Molar mass of [tex]NaOH[/tex] = 40 g/mol
First we have to calculate the moles of [tex]H_2SO_4[/tex] and [tex]NaOH[/tex].
[tex]\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}[/tex]
[tex]\text{Moles of }H_2SO_4=\frac{3.9g}{98g/mol}=0.0398mol[/tex]
and,
[tex]\text{Moles of }NaOH=\frac{\text{Given mass }NaOH}{\text{Molar mass }NaOH}[/tex]
[tex]\text{Moles of }NaOH=\frac{7.8g}{40g/mol}=0.195mol[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical equation is:
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]H_2SO_4[/tex] react with 2 mole of [tex]NaOH[/tex]
So, 0.0398 moles of [tex]H_2SO_4[/tex] react with [tex]0.0398\times 2=0.0796[/tex] moles of [tex]NaOH[/tex]
From this we conclude that, [tex]NaOH[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]H_2SO_4[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]Na_2SO_4[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]H_2SO_4[/tex] react to give 1 mole of [tex]Na_2SO_4[/tex]
So, 0.0398 mole of [tex]H_2SO_4[/tex] react to give 0.0398 mole of [tex]Na_2SO_4[/tex]
Now we have to calculate the mass of [tex]Na_2SO_4[/tex]
[tex]\text{ Mass of }Na_2SO_4=\text{ Moles of }Na_2SO_4\times \text{ Molar mass of }Na_2SO_4[/tex]
Molar mass of [tex]Na_2SO_4[/tex] = 142 g/mole
[tex]\text{ Mass of }Na_2SO_4=(0.0398moles)\times (142g/mole)=5.65g[/tex]
Therefore, the theoretical yield of [tex]Na_2SO_4[/tex] is, 5.65 grams.
