Question

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

Answer 1

The given question is incomplete. The complete question is:

Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O). If 1.60 g of sodium chloride is produced from the reaction of 1.8 g of hydrochloric acid and 1.4 g of sodium hydroxide, calculate the percent yield of sodium chloride. Be sure your answer has the correct number of significant digits in it.

Answer: Thus the percent yield of sodium chloride is 78.0%

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}*{\text{Molar Mass}}`

`\text{Moles of} HCl=(1.8g)/(36.5g/mol)=0.049moles`

`\text{Moles of} NaOH=(1.4g)/(40g/mol)=0.035moles`

`HCl(aq)+NaOH(aq)\rightarrow NaCl(aq)+H_2O(l)`

According to stoichiometry :

1 mole of
`NaOH` require = 1 mole of
`HCl`

Thus 0.035 moles of
`NaOH` will require=
`(1)/(1)* 0.035=0.035moles` of
`HCl`

Thus
`NaOH` is the limiting reagent as it limits the formation of product and
`HCl` is the excess reagent.

As 1 mole of
`NaOH` give = 1 mole of
`NaCl`

Thus 0.035 moles of
`NaOH` give =
`(1)/(1)* 0.035=0.035moles` of
`NaCl`

Mass of
`NaCl=moles* {\text {Molar mass}}=0.035moles* 58.5g/mol=2.05g`

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

`{\text {percentage yield}}=(1.60g)/(2.05g)* 100\%=78.0\%`

Thus the percent yield of sodium chloride is 78.0%