Question

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?---------

Explanation please

Answer 1

You have the balanced equations.
2. Convert 157 g NaCl to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles NaCl to moles Na2CO3.
4. Now convert moles Na2CO3 to grams. g = moles x molar mass. This the theoretical yield.
5. %yield = (actual yield/theoretical yield)*100 = ??

The problem gives 93.8 g Na2CO3 as the actual yield.