Problem PageQuestion Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . What is the theoretical yield of water formed from the reaction of of hydrochloric acid and of sodium hydroxide

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.

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%