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 159 g of NaCl were used in the first reaction? If 78.8 g of Na2CO3 was obtained, what was the percentage yield?
The reactions involved are
H2O + NaCl + NH3 +CO2 ---------> NH4Cl + NaHCO3
2 NaHCO3 -------------------------------------> Na2CO3 +CO2 + H2O
Using the principle of atomic conservation for Na, that is assuming all the Na is converted completely into the final product.
Thus moles of Na in NaCl = moles of na in product
159/58.5 = 2 x weight of Na2CO3 / 106
Thus weight of Na2CO3 otained theoretically = 143.15
If 78.8g of Na2CO3 is obtained then % yield = Na2CO3 obtianedx100 / theoretical yield
= [78.8x100] /143.15
= 55.05%
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