Calculate the percent yield of NaCl if 1.763 g of NaCl are produced from the reaction of 2.653 g of Na2CO3 with 125 mL of 0.317M HCl according to the following reaction. The M stands for Molarity.
Na2CO3 + 2 HCl yields 2 NaCl + CO2 + H2O
I have done this problem and I coming out with a number that is higher than the theoretical yield.
I really appreciate the help, and thank you in advance!
Na2CO3 + 2 HCl yields 2 NaCl + CO2 + H2O
I have done this problem and I coming out with a number that is higher than the theoretical yield.
I really appreciate the help, and thank you in advance!
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Na2CO3 + 2 HCl → 2 NaCl + CO2 + H2O
(2.653 g Na2CO3) / (105.9886 g/mol) = 0.02503 mole Na2CO3
(0.125 L) x (0.317 mol/L) = 0.039625 mole HCl
0.039625 mole of HCl would react completely with 0.039625 / 2 = 0.0198125 mole of Na2CO3, but there is more Na2CO3 present than that, so Na2CO3 is in excess and HCl is the limiting reactant.
(0.039625 mol HCl) x (2/2) x (58.4430 g NaCl/mol) = 2.3158 g NaCl in theory
1.763 g / 2.3158 g = 0.76129 = 76.13%
(2.653 g Na2CO3) / (105.9886 g/mol) = 0.02503 mole Na2CO3
(0.125 L) x (0.317 mol/L) = 0.039625 mole HCl
0.039625 mole of HCl would react completely with 0.039625 / 2 = 0.0198125 mole of Na2CO3, but there is more Na2CO3 present than that, so Na2CO3 is in excess and HCl is the limiting reactant.
(0.039625 mol HCl) x (2/2) x (58.4430 g NaCl/mol) = 2.3158 g NaCl in theory
1.763 g / 2.3158 g = 0.76129 = 76.13%