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. Be sure your answer has the correct number of significant digits in it.

Answer 1

Answer: The percent yield of water is 46.9 %

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For sulfuric acid:

Given mass of sulfuric acid = 72.6 g (Assuming)

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 1, we get:

`\text{Moles of sulfuric acid}=(72.6g)/(98g/mol)=0.741mol`

• For NaOH:

Given mass of NaOH = 77 g (Assuming)

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

`\text{Moles of NaOH}=(77g)/(40g/mol)=1.925mol`

The chemical equation for the reaction of sulfuric acid and sodium hydroxide follows:

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

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of NaOH

0.741 moles of sulfuric acid will react with =
`(2)/(1)* 0.741=1.482mol` of NaOH

As, given amount of NaOH is more than the required amount. So, it is considered as an excess reagent.

Thus, sulfuric acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sulfuric acid reacts with 2 moles of water

0.741 moles of sulfuric acid will react with =
`(2)/(1)* 0.741=1.482mol` of water

• Now, calculating the mass of water from equation 1, we get:

Molar mass of water = 18 g/mol

Moles of water = 1.482 moles

Putting values in equation 1, we get:

`1.482mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(1.482mol* 18g/mol)=26.67g`

• To calculate the percentage yield of water, we use the equation:

`\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}* 100`

Experimental yield of water = 12.5 g (Assuming)

Theoretical yield of water = 26.67 g

Putting values in above equation, we get:

`\%\text{ yield of water}=(12.5g)/(26.67g)* 100\\\\\% \text{yield of water}=46.9\%`

Hence, the percent yield of water is 46.9 %