Answer:
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 %
