To determine the theoretical yield of sodium carbonate in grams, you need to use stoichiometry and the balanced equation of the reaction.
First, we need to calculate the molar mass of NaCl. The molar mass of Na is 22.99 g/mol, and the molar mass of Cl is 35.45 g/mol. So the molar mass of NaCl is:
(22.99 g/mol Na) + (35.45 g/mol Cl) = 58.44 g/mol NaCl
Next, we need to convert the mass of NaCl used to moles. We can do this by dividing the given mass by the molar mass:
157 g NaCl / 58.44 g/mol NaCl = 2.69 mol NaCl
Now, let's examine the balanced equation to determine the stoichiometry between NaCl and Na2CO3. From the equation, we see that it takes 1 mole of NaCl to produce 1 mole of Na2CO3. Therefore, the moles of Na2CO3 produced are also 2.69 mol.
Finally, we can calculate the mass of Na2CO3 using the molar mass of Na2CO3. The molar mass of Na2CO3 is:
(22.99 g/mol Na * 2) + (12.01 g/mol C) + (16.00 g/mol O * 3) = 105.99 g/mol Na2CO3
So, the theoretical yield of Na2CO3 is:
2.69 mol Na2CO3 * 105.99 g/mol Na2CO3 = 285 g Na2CO3
Therefore, the theoretical yield of sodium carbonate is 285 grams.
To calculate the percentage yield, we need to use the actual yield and divide it by the theoretical yield, then multiply by 100.
Given: Actual yield = 93.8 g Na2CO3
Percentage yield = (Actual yield / Theoretical yield) * 100
= (93.8 g / 285 g) * 100
= 32.91%
Therefore, the percentage yield of sodium carbonate is approximately 32.91%.