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.

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

87.9 % is the percent yield of H₂O

Explanation:

This is the neutralization reaction. A base reacts with an acid to produce water and the correspondly ionic salt.

NaOH  +  HCl  → NaCl +  H₂O

As we have the mass of the two reactants, we must determine the limiting reactant.

Let's convert to moles, the mass of each reactant. (mass / molar mass)

21.1 g / 36.45 g/mol = 0.579 moles of HCl

46.3 g / 40g/mol = 1.15 moles of NaOH

Ratio is 1:1, so it is obviously that the limiting reactant is the HCl. For 1.15 moles of NaOH, i need the same amount of acid, but I only have 0.579 moles

Let's work with the products now. Ratio is 1:1 again, so If I have 0.579 moles of acid, I can produce 0.579 moles of H₂O.

How many grams are 0.579 moles of water? We should find it out as this

mol . molar mass = mass → 0.579 mol . 18 g/mol = 10.4 g

We were told that the production of water was 9.17 g, so let's determine the percent yield as this:

(Yield produced / Theoretical yield) . 100 =

(9.17 g / 10.4g ) . 100 = 87.9 %