Question

Aqueous hydrochloric acid reacts with solid sodium hydroxide to produce aqueous sodium chloride and liquid water . If of water is produced from the reaction of of hydrochloric acid and of sodium hydroxide, calculate the percent yield of water. Round your answer to significant figures.

Answer 1

Answer:

87.9%

Explanation:

Balanced Chemical Equation:

HCl + NaOH = NaCl + H2O

We are Given:

Mass of H2O = 9.17 g

Mass of HCl = 21.1 g

Mass of NaOH = 43.6 g

First, calculate the moles of both HCl and NaOH:

Moles of HCl: 21.1 g of HCl x 1 mole of HCl/36.46 g of HCl = 0.579 moles

Moles of NaOH: 43.6 g of NaOH x 1 mole of NaOH/40.00 g of NaOH = 1.09 moles

Here you calculate the mole of H2O from the moles of both HCl and NaOH using the balanced chemical equation:

Moles of H2O from the moles of HCl: 0.579 moles of HCl x 1 mole of H2O/1 mole of HCl = 0.579 moles

Moles of H2O from the moles of NaOH: 1.09 moles of HCl x 1 mole of H2O/1 mole of NaOH = 1.09 moles

From the calculations above, we can see that the limiting reagent is HCl because it produced the lower amount of moles of H2O. Therefore, we use 0.579 moles and NOT 1.09 moles to calculate the mass of H2O:

Mass of H2O: 0.579 moles of H2O x 18.02 g of H2O/1 mole of H2O = 10.43 g

% yield of H2O = actual yield/theoretical yield x 100= 9.17 g/10.43 g x 100 = 87.9%