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=(72.6g)/(98g/mol)=0.741mol`

and,

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

`\text{Moles of }NaOH=(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* 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* 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* \text{ Molar mass of }H_2O`

Molar mass of
`H_2O` = 18 g/mole

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

Now we have to calculate the percent yield of water.

`\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}* 100`

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

`\text{Percent yield}=(12.5g)/(26.68g)* 100=46.8\%`

Thus, the percent yield of water is, 46.8 %