Respuesta :
This is an incomplete question, here is a complete question.
Aqueous sulfuric acid (H₂SO₄) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium sulfate (Na₂SO₄) and liquid water (H₂O) . If 12.5 g of water is produced from the reaction of 72.6 g of sulfuric acid and 77.0 g of sodium hydroxide, calculate the percent yield of water. Round your answer in significant figures.
Answer: The percent yield of water is, 46.8 %
Explanation : Given,
Mass of [tex]H_2SO_4[/tex] = 72.6 g
Mass of [tex]NaOH[/tex] = 77.0 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{72.6g}{98g/mol}=0.741mol[/tex]
and,
[tex]\text{Moles of }NaOH=\frac{\text{Given mass }NaOH}{\text{Molar mass }NaOH}[/tex]
[tex]\text{Moles of }NaOH=\frac{77.0g}{40g/mol}=1.925mol[/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, 0.741 moles of [tex]H_2SO_4[/tex] react with [tex]0.741\times 2=1.482[/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]H_2O[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]H_2SO_4[/tex] react to give 2 mole of [tex]H_2O[/tex]
So, 0.741 moles of [tex]H_2SO_4[/tex] react to give [tex]0.741\times 2=1.482[/tex] moles of [tex]H_2O[/tex]
Now we have to calculate the mass of [tex]H_2O[/tex]
[tex]\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O[/tex]
Molar mass of [tex]H_2O[/tex] = 18 g/mole
[tex]\text{ Mass of }H_2O=(1.482moles)\times (18g/mole)=26.68g[/tex]
Now we have to calculate the percent yield of water.
[tex]\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Percent yield}=\frac{12.5g}{26.68g}\times 100=46.8\%[/tex]
Thus, the percent yield of water is, 46.8 %
