Question

Aqueous sulfuric acid reacts with solid sodium hydroxide to produce aqueous sodium sulfate and liquid water . If of water is produced from the reaction of of sulfuric acid and of sodium hydroxide, calculate the percent yield of water. Round your answer to significant figures.

Answer 1

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 $H_2SO_4$ = 72.6 g

Mass of $NaOH$ = 77.0 g

Molar mass of $H_2SO_4$ = 98 g/mol

Molar mass of $NaOH$ = 40 g/mol

First we have to calculate the moles of $H_2SO_4$ and $NaOH$.

$\text{Moles of }H_2SO_4=\frac{\text{Given mass }H_2SO_4}{\text{Molar mass }H_2SO_4}$

$\text{Moles of }H_2SO_4=\frac{72.6g}{98g/mol}=0.741mol$

and,

$\text{Moles of }NaOH=\frac{\text{Given mass }NaOH}{\text{Molar mass }NaOH}$

$\text{Moles of }NaOH=\frac{77.0g}{40g/mol}=1.925mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $H_2SO_4$ react with 2 mole of $NaOH$

So, 0.741 moles of $H_2SO_4$ react with $0.741\times 2=1.482$ moles of $NaOH$

From this we conclude that, $NaOH$ is an excess reagent because the given moles are greater than the required moles and $H_2SO_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 1 mole of $H_2SO_4$ react to give 2 mole of $H_2O$

So, 0.741 moles of $H_2SO_4$ react to give $0.741\times 2=1.482$ moles of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

Molar mass of $H_2O$ = 18 g/mole

$\text{ Mass of }H_2O=(1.482moles)\times (18g/mole)=26.68g$

Now we have to calculate the percent yield of water.

$\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

Now put all the given values in this formula, we get:

$\text{Percent yield}=\frac{12.5g}{26.68g}\times 100=46.8\%$

Thus, the percent yield of water is, 46.8 %