Question

Aqueous hydrochloric acid (HCl) reacts with solid hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water H2O. If 1.23 g of sodium chloride is produced from the reaction of 2.2 g of hydrochloric acid and 4.6 g of sodium hydroxide, calculate the percent yield of sodium chloride. Round your answer to 2 significant digits.

Answer 1

First we will write the balanced chemical reaction for this reaction

NaOH_(aqueous) + HCl_(aqueous) ---------> NaCl_(aqueous) + H₂O_(l)

This reaction shows 1:1 ratio I.e. 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and H₂O this shows that  this reaction theoretically has a 100% yield. But it is not possible in actual practice so we will find the mole used in our reaction using a simple formula

No. Of moles = mass / molar mass

1. NaOH

Mass = 4.6 g molar mass = 40

No. Of moles of NaOH = 4.6/40 = 0.115 g / mol

2. HCl

Mass = 2.2 g molar mass = 36

No. Of moles of HCl = 2.2/ 36 = 0.06 g/ mol

3. NaCl

Mass = 1.23 g molar mass = 58

No. Of moles of NaCl = 1.23/ 58 = 0.021g/ mol

So you will use a basic formula

% yield = (0.021/0.06) * 100

% yield= 35%