percent yield of water

Question

Aqueous hydrochloric acid HCl reacts with solid sodium hydroxide NaOH to produce aqueous sodium chloride NaCl and liquid water H2O.

If 0.321g of water is produced from the reaction of 2.9g of hydrochloric acid and 0.83g of sodium hydroxide, calculate the percent yield of water.

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

Answer 1

HCl(aq) + NaOH(s) ==> NaCl(aq) + H2O(l) ... balanced equation

2.9 g.......0.83 g...............................0.321 g.........

We will first find if either reactant is in limiting supply:

moles HCl present = 2.9 g x 1 mol HCl/36.5 g = 0.0795 moles HCl

moles NaOH present = 0.83 g x 1 mol NaOH/40 g = 0.0208 moles NaOH

Since the balanced equation tells us that these reagents react in a 1:1 mole ratio, NaOH is LIMITING

Theoretical yield will be based on moles of NaOH since it is limiting.

Theoretical yield = 0.0208 moles NaOH x 1 mol H2O/mole NaOH x 18 g H2O/mol H2O = 0.374 g H2O

Percent yield = actual yield/theoretical yield (x100) = 0.321 g / 0.374 g (x100) = 85.8% yield