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%

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}\times{\text{Molar Mass}}$    

$\text{Moles of} HCl=\frac{1.8g}{36.5g/mol}=0.049moles$

$\text{Moles of} NaOH=\frac{1.4g}{40g/mol}=0.035moles$

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

According to stoichiometry :

1 mole of $NaOH$ require = 1 mole of $HCl$

Thus 0.035 moles of $NaOH$ will require=$\frac{1}{1}\times 0.035=0.035moles$ 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 =$\frac{1}{1}\times 0.035=0.035moles$  of $NaCl$

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

${\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%$

${\text {percentage yield}}=\frac{1.60g}{2.05g}\times 100\%=78.0\%$

Thus the percent yield of sodium chloride is 78.0%