Please help
2P+5CI2=>2PCI5
.473g P and 3.25 CIs
Actual yield=2.12g PCI5
Find the percent yield. Please show work.
2P+5CI2=>2PCI5
.473g P and 3.25 CIs
Actual yield=2.12g PCI5
Find the percent yield. Please show work.
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2 P + 5 Cl2 → 2 PCl5
(0.473 g P) / (30.97376 g P/mol) = 0.015271 mol P
Supposing the missing units to be grams:
(3.25 g Cl2) / (70.9064 g Cl2/mol) = 0.045835 mol Cl2
0.015271 mole of P would react completely with 0.015271 x (5/2) = 0.0381775 mole Cl2, but there is more Cl2 present than that, so Cl2 is in excess and P is the limiting reactant.
(0.015271 mol P) x (2/2) x (208.2398 g PCl5/mol) = 3.18 g PCl5 in theory
(2.12 g) / (3.18 g) = 0.667 = 66.7%
(0.473 g P) / (30.97376 g P/mol) = 0.015271 mol P
Supposing the missing units to be grams:
(3.25 g Cl2) / (70.9064 g Cl2/mol) = 0.045835 mol Cl2
0.015271 mole of P would react completely with 0.015271 x (5/2) = 0.0381775 mole Cl2, but there is more Cl2 present than that, so Cl2 is in excess and P is the limiting reactant.
(0.015271 mol P) x (2/2) x (208.2398 g PCl5/mol) = 3.18 g PCl5 in theory
(2.12 g) / (3.18 g) = 0.667 = 66.7%