Answer to Question #231792 in General Chemistry for Tim

Question

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O. If 0.438g of sodium chloride is produced from the reaction of 2.9g of hydrochloric acid and 1.2g of sodium hydroxide, calculate the percent yield of sodium chloride.

Round your answer to 2 significant figures.

Answer 1

$HCl + NaOH \to NaCl + H_2O$

no. of moles of HCl in the reaction = 2.9/36.5 = 0.08 moles

no. of moles of NaOH in the reaction = 1.2/40 = 0.03 moles

$\therefore$ the limiting reagent is NaOH

no. of moles of NaCl produced = 0.438/58.5 = 0.0075 moles

no. of moles of NaCl that should have been produced = 0.03 moles

$\%\textsf{yield}=\dfrac{0.075}{0.03}× 100\% = 2.5\%$