Final answer:
The theoretical yield for the reaction of 30g HCl with excess NaOH would be 48.08g of NaCl based on stoichiometry and molecular weight calculations. Since only 10g of NaCl was actually produced, the percent yield of the experiment is 20.80%.
Step-by-step explanation:
Theoretical Yield Calculation
To determine the theoretical yield of an experiment, we must first understand the balanced chemical equation of the reaction. For the reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), the balanced equation is:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
This equation shows that one mole of HCl reacts with one mole of NaOH to produce one mole of NaCl and one mole of water. If a student reacts 30g of HCl with excess NaOH and produces 10g of NaCl, we would calculate the theoretical yield by first determining the moles of each compound based on their molecular weights. We then use the mole ratio from the balanced equation to find the maximal amount of product that can form.
The molecular weight of HCl is approximately 36.46 g/mol, so the moles of HCl used are:
30g HCl × (1 mol HCl / 36.46 g HCl) = 0.8229 mol HCl
Given the 1:1 mole ratio in the reaction, the theoretical yield of NaCl would also be 0.8229 mol, since no limit is given for NaOH. We then convert moles of NaCl to grams using its molecular weight (58.44 g/mol):
0.8229 mol NaCl × (58.44 g NaCl / 1 mol NaCl) = 48.08 g NaCl (theoretical yield)
However, since only 10g of NaCl was actually produced, this is the actual yield. The percent yield can be calculated by dividing the actual yield by the theoretical yield and then multiplying by 100:
(10g / 48.08g) × 100 = 20.80%