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.

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%