If 49.0g of H3PO4 is reacted with excess KOH, determine the percent yield of K3PO4 if you isolate 49.0g of K3PO4

Question

Given the the equation : H3PO4 + KOH

Answer 1

First, write the balance equation:

H₃PO₄ + 3 KOH ==> K₃PO₄ + 3 H₂O

To find % yield, we first must find the theoretical yield:

49.0 g H₃PO₄ x 1 mol H₃PO₄/97.99 g = 0.500 moles H₃PO₄ used

0.500 mol H₃PO₄ x 1 mol K₃PO₄/1 mol H₃PO₄ x 212 g K₃PO₄/mole = 106 g K₃PO₄ = theoretical yld

% yield = actual yield/theoretical yield (x100%) = 49.0 g/106 g (x100%) = 46.2% yield

Answer 2

This is a stoichiometry problem at heart.

Also notice that this is a neutralization reaction so the student is expected to know what will be the products

First step is to balance the equation

H3PO4 + 3KOH ---> K3PO4 + 3H2O

Second step is to convert values to moles. In this case since you are given the amount that will be produced, hang on to that for finding the percent yield. Remember that %yield =( Actual/Theoretical) X 100

49 g H3PO4 can be converted to moles by dividing by the Formula Mass or Gram Molecular Mass. The formula Mass for H3PO4 is 97.944

Therefore 49/97.944 = 0.500 Moles

Third Step is to use the molar ratio to discover the number of moles that should theoretically be produced

In this case since the molar ratio is 1:1 (numbers in front of the formulas in the balanced equation) 0.500 moles of K3PO4 should be produced. This is the theoretical yield

Fourth step is to convert theoretical moles to grams by multiplying the theoretical moles produced (0.500 moles) by the Formula Mass of K3PO4. This mass is 212.264.

Therefore 0.0500 X 212.264 = 106.132 g This should be the theoretical grams produced

If 49.0 grams were recovered

Then (49.0/106.132) X 100 is the percent yield

46.169 is the percent yield Rounded to the correct number of significant figures your answer should be 46.2 %