Aqueous hydrochloric acid (HC1) reacts with solid sodium hydroxide (NaOH) to produce aqueous sodium chloride (NaC1) and liquid water (H2O). What is the theoretical yield of water formed from the reaction of 29.5 g of hydrochloric acid and 58.0 g of sodium hydroxide?
Round your answer to 3 significant figures.
First, we need to write a balanced chemical equation for the reaction:
HCl + NaOH -> NaCl + H2O
Next, we need to determine the limiting reactant.
1. Calculate the number of moles of each reactant:
- Moles of HCl = 29.5 g / 36.46 g/mol = 0.808 mol
- Moles of NaOH = 58.0 g / 40.00 g/mol = 1.45 mol
2. Determine the mole ratio of the reactants:
- From the balanced equation, the mole ratio of HCl to NaOH is 1:1.
3. Since the mole ratio is 1:1, HCl is the limiting reactant because it produces fewer moles of product.
4. Calculate the theoretical yield of water using the moles of the limiting reactant:
- 0.808 mol HCl * 1 mol H2O / 1 mol HCl = 0.808 mol H2O
5. Convert the moles of water to grams:
- 0.808 mol H2O * 18.015 g/mol = 14.6 g H2O
Therefore, the theoretical yield of water formed from the reaction of 29.5 g of hydrochloric acid and 58.0 g of sodium hydroxide is 14.6 g.