The Solvay process for the manufacture of sodium carbonate begins by passing ammonia and carbon dioxide through a solution of sodium chloride to make sodium bicarbonate and ammonium chloride. The equation for this reaction is H2O + NaCl + NH3 + CO2 �¨ NH4Cl + NaHCO3. In the next step, sodium bicarbonate is heated to give sodium carbonate and two gases, carbon dioxide and steam 2NaHCO3 �¨ Na2CO3 + CO2 + H2O.

What is the theoretical yield of sodium carbonate, expressed in grams, if 157 g of NaCl were used in the first reaction?---------

If 93.8 g of Na2CO3 were obtained from the reaction described in part a, what was the percentage yield?---------

please write the anwer and i will work it out

I have a better idea. You work it out and post it and I'll check it. Post your work, also, if you want me to look for errors or just the answer if you want to know if you have it right.

To determine the theoretical yield of sodium carbonate, we need to use stoichiometry and the molar ratios provided in the balanced chemical equation.

Given: Mass of NaCl = 157 g

1. Convert the mass of NaCl to moles:

Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

Number of moles of NaCl = Mass of NaCl / Molar mass of NaCl
= 157 g / 58.44 g/mol
= 2.69 mol

2. Use the stoichiometric coefficients from the balanced equation to find the moles of Na2CO3 formed:

From the equation: 2NaHCO3 -> Na2CO3 + CO2 + H2O

The molar ratio between NaCl and Na2CO3 is 1:1. Therefore, the moles of Na2CO3 formed will be the same as the moles of NaCl used, which is 2.69 mol.

3. Calculate the molar mass of Na2CO3:

Molar mass of Na2CO3 = 22.99 g/mol + 2(12.01 g/mol) + 3(16.00 g/mol)
= 105.99 g/mol

4. Calculate the theoretical yield of Na2CO3:

Theoretical yield of Na2CO3 = Moles of Na2CO3 * Molar mass of Na2CO3
= 2.69 mol * 105.99 g/mol
≈ 285.9 g

Therefore, the theoretical yield of sodium carbonate is approximately 285.9 grams when 157 grams of NaCl are used in the first reaction.

To find the percentage yield:

Given: Mass of Na2CO3 obtained = 93.8 g

Percentage yield = (Mass of obtained Na2CO3 / Theoretical yield of Na2CO3) * 100
= (93.8 g / 285.9 g) * 100
≈ 32.8%

Therefore, the percentage yield of sodium carbonate is approximately 32.8%.