In industry, sodium carbonate is produced in two stages whose equations are as follows:
NaCl + NH3 + H2O + CO2 à NH4Cl + NaHCO3
2NaHCO3 + heat à Na2CO3 + CO2 + H2O
a) What’s the mass of sodium carbonate that can be produced from 1.0 t of sodium chloride? (Assume that the other reactants are in excess.)
Hint: Indicate clearly how you found the molar ratio. Use the two equations given to you above to make your calculations.
b) If 0.85 t of sodium carbonate is actually produced from 1.0 t of sodium chloride, what’s the theoretical yield percentage?
To find the mass of sodium carbonate that can be produced from 1.0 t of sodium chloride, we need to use the given equations and determine the molar ratio between sodium chloride (NaCl) and sodium carbonate (Na2CO3).
a) First, let's determine the molar ratio between NaCl and Na2CO3 using the given equations:
From the first equation:
1 mole of NaCl produces 1 mole of NaHCO3
From the second equation:
2 moles of NaHCO3 produce 1 mole of Na2CO3
Combining the two equations, we can see that:
1 mole of NaCl produces 1 mole of NaHCO3, which produces 1/2 mole of Na2CO3
From the molar mass of NaCl (58.44 g/mol) and Na2CO3 (105.99 g/mol), we can calculate the molar mass ratio:
58.44 g of NaCl produces 105.99 g of Na2CO3
Now, let's calculate the mass of Na2CO3 that can be produced from 1.0 t (1000 kg) of NaCl:
(1000 kg) x (1000 g/kg) / (58.44 g/mol) x (105.99 g/mol) = 29310.8 g
Therefore, the mass of sodium carbonate that can be produced from 1.0 t of sodium chloride is approximately 29310.8 g, or 29.31 kg.
b) To calculate the theoretical yield percentage, we need to compare the actual yield (0.85 t or 850 kg) to the theoretical yield (29.31 kg) calculated in part a:
Theoretical yield percentage = (Actual yield / Theoretical yield) x 100
= (850 kg / 29.31 kg) x 100
= 2904.3%
Therefore, the theoretical yield percentage is approximately 2904.3%.