Aqueous hydrochloric acid (HCI) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaCl) and liquid water (H20). What is the theoretical yield of sodium chloride formed from the reaction of 11.7 g of hydrochloric acid and 22.4 g of sodium hydroxide?

Question

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

Answer 1

First, we need to determine the limiting reactant in the reaction.

1. Calculate the molar mass of each compound:
- HCl: 1(1.008) + 1(35.453) = 36.461 g/mol
- NaOH: 22.99 + 1.008 + 15.999 = 39.997 g/mol

2. Calculate the moles of each reactant:
- Moles of HCl = 11.7 g / 36.461 g/mol = 0.320 moles
- Moles of NaOH = 22.4 g / 39.997 g/mol = 0.560 moles

3. Determine the limiting reactant (HCl in this case).

4. Use the mole ratio from the balanced chemical equation to determine the moles of NaCl produced:
2 moles HCl : 2 moles NaCl

Moles of NaCl = 0.320 moles HCl * (2 moles NaCl / 2 moles HCl) = 0.320 moles NaCl

5. Calculate the mass of NaCl produced:
Mass of NaCl = 0.320 moles * (22.99 + 35.453) = 19.24 g

Therefore, the theoretical yield of sodium chloride formed from the reaction of 11.7 g of hydrochloric acid and 22.4 g of sodium hydroxide is 19.24 g.