Question

3. Answer the following with this reaction: P4O10+6PCl5 = 10POCl3

a. How many grams of POCl3 are produced when 225.0 grams of P4O10 and 675.0 grams of PCl5 react?

b. How many grams of excess reactant remain after the reaction in (a)?

c. When 42.66 grams of PCl5 react with excess P4O10, the amount of product formed is 47.22 grams of POCl3. What is the percent yield?

Answer 1

P4O10 + 6PCl5 ------------> 10POCl3

a) 283.886 g P4O10 reacts with 6 x 208.24 g PCl5

X g P4O10 reacts with 675.0 g PCl5

X = 675 x 283.886 / 6 x 208.24 = 153.4 g P4O10

we have 225.0 g P4O10 so P4O10 is exess reagent.

PCl5 is limiting reagent.

6 x 208.24 g PCl5 gives 10 x 153.3 g POCl3

675.0 g PCl5 gives 675 x 10 x 153.3 / 6 x 208.24 = 828.2 g

mass of POCl3 formed = 828.2 g

b) exess P4O10 remains = 225.0 - 153.4 = 71.6 g

c) 6 x 208.24 g PCl5 gives 10 x 153.3 g POCl3

42.66 g PCl5 gives 42.66 x 10 x 153.3 / 6 x 208.24 = 52.34 g POCL3

actual yield = 47.22 g

% yield = (47.22 / 52.34) x 100

% yield = 90.2