complete question:
Aqueous hydrochloric acid (HCl) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H2O) . If 9.17g of water is produced from the reaction of 21.1g of hydrochloric acid and 43.6 of sodium hydroxide, calculate the percent yield of water.
Answer:
percentage yield of water = 88.10%
Explanation:
Firstly, let us write the chemical equation of the reaction and balance it.
HCl + NaOH → NaCl + H2O
we need to know the limiting reactant by converting the reactants to moles
molar mass of HCl = 1 + 35.5 = 36.5 g
moles of HCl = mass/molar mass = 21.1/36.5 = 0.57808219178 mol-rxn
molar mass of NaOH = 23 + 16 + 1 = 40 g
moles of NaOH = 43.6/40 = 1.09 mol-rxn
The limiting reactant is HCl and it will determine the amount of water produced.
Molar mass of H2O = 2 + 16 = 18 g
if 36.5 g of HCl produces 18 g of H2O
21.1 g of HCl will produce ? grams of H2O
cross multiply
theoretical yield of H2O = (21.1 × 18)/36.5
theoretical yield of H2O =379.8
/36.5
theoretical yield of H2O = 10.4054794521
theoretical yield of H2O = 10.41 g
percentage yield of water = actual yield/ theoretical yield × 100
percentage yield of water = 9.17/10.41 × 100
percentage yield of water = 917/10.41
percentage yield of water = 88.088376561
percentage yield of water = 88.10%
