Question

What is the percentage yield of a reaction in which 200g pcl3 reacts with excess water to form 127g of hcl

Answer 1

79.7 %

Step-by-step explanation:

PCl3 + 3H2O → H3PO3 + 3HCl
1 mole of PCl3 reacts with 3 moles of water to produce 1 mole of H3PO3 and 3 moles of HCl.

Given, 200g of PCl3 (amu = 137.5)
which is 200/137.5 moles of PCl3

Now, if we need 1 mole of PCl3 to produce 3 moles of HCl
then 200/137.5 moles of PCl3 would produce
(200 * 3)/ 137.5 = 4.36 moles of HCl
which weighs 159.27 g of HCl

But we only managed to yield 127 g of HCl

So the percentage yield would be (127/159.27) * 100 = 79.7%