Final answer:
The theoretical yield of sodium carbonate from the Solvay process when 120 g of NaCl is used is approximately 108.8 g. The actual yield obtained was 85.4 g, resulting in a percentage yield of 78.49%. None of the options provided in the question accurately represents the theoretical yield as calculated.
Step-by-step explanation:
To determine the theoretical yield of sodium carbonate from the Solvay process, we must consider the stoichiometry of the given reactions. First, we need to convert the mass of sodium chloride (NaCl) to moles using its molar mass (58.44 g/mol). For 120 g of NaCl:
moles of NaCl = 120 g / 58.44 g/mol = 2.053 moles
Next, according to the reaction stoichiometry, sodium bicarbonate (NaHCO3) will be formed, which will then decompose to sodium carbonate (Na2CO3), following these equations:
2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(l)
From the stoichiometry, it's clear that two moles of NaHCO3 yields one mole of Na2CO3. Therefore, the moles of Na2CO3 formed would be half of the moles of NaCl used:
moles of Na2CO3 = 2.053 moles NaCl / 2 = 1.0265 moles Na2CO3
To convert moles of Na2CO3 to grams, we use its molar mass (105.99 g/mol):
theoretical yield of Na2CO3 = 1.0265 moles * 105.99 g/mol = 108.8 g
To find the percentage yield, we use the actual yield and the theoretical yield:
percentage yield = (actual yield / theoretical yield) * 100
percentage yield = (85.4 g / 108.8 g) * 100 = 78.49%
The correct theoretical yield of sodium carbonate is close to c. Theoretical yield = 108.8 g (rounded to the nearest gram, c. 109 g could be an option if provided), but as per the given options, none exactly matches the calculated value. If we must choose one of the given options, none are correct. However, the closest numerical value provided is option c. Theoretical yield = 200 g, which is still not accurate.