Final answer:
The theoretical yield of NaHCO3 from the Solvay experiment is 21.0025 grams based on the 0.250 moles of CO2 as the limiting reactant. The difference between the theoretical and experimental yield (which was 7.2340 grams) could be due to incomplete reactions, product recovery losses, reactant impurities, or measurement errors.
Step-by-step explanation:
To calculate the theoretical yield of NaHCO3 from the Solvay experiment, we need to first determine the limiting reactant. For this reaction, we are given 40.0 ml of 4.00 M ammonia solution, which amounts to 0.160 moles of NH3 (since Molarity (M) is moles of solute per liter of solution, and 40.0 ml is 0.040 liters). Also provided is 17.0 grams of NaCl and 0.250 moles of CO2. Without the full equation of the Solvay process, we assume those are the reactants that produce NaHCO3.
First, we must convert the mass of NaCl to moles: 17.0 g of NaCl (molar mass = 58.44 g/mol) yields 0.291 moles of NaCl. The next step is determining the limiting reactant by comparing the stoichiometric ratios in the balanced equation of the process, which typically is as follows: 2 NH3 + 2 CO2 + 2 NaCl → 2 NaHCO3 + 2 NH4Cl. From this, we can see that the limiting reactant is CO2 because we have less than the required stoichiometric amount as compared to NH3 and NaCl.
The theoretical yield of NaHCO3 would thus be equivalent to the moles of CO2 since they react in a 1:1 ratio in the equation. So, we expect 0.250 moles of NaHCO3, and since the molar mass of NaHCO3 is 84.01 g/mol, the theoretical yield is 0.250 moles * 84.01 g/mol = 21.0025 grams of NaHCO3.
The difference between the theoretical yield and experimental yield of the NaHCO3 precipitate (which is 7.2340 grams) could be because of several factors, including incomplete reaction, losses during product recovery, impurities in reactants, or measurement errors.