Question

Consider the following balanced equation:

2H3PO4(aq) + 3Ca(OH)2(s) → 6H2O(l) + Ca3(PO4)2(s)

If 13.6 moles of H3PO4(aq) and 35.3 moles of Ca(OH)2(s) are allowed to react, and the percent yield is 84.5%, how many moles of H2O(l) will actually be produced?

60.5 moles

87.2 moles

95.2 moles

76.7 moles

34.5 moles

Answer 1

Balanced chemical equation is:

2 H3PO4 + 3 Ca(OH)2 ---> 6 H2O + Ca3(PO4)2

2 mol of H3PO4 reacts with 3 mol of Ca(OH)2

for 13.6 mol of H3PO4, 20.4 mol of Ca(OH)2 is required

But we have 35.3 mol of Ca(OH)2

so, H3PO4 is limiting reagent

we will use H3PO4 in further calculation

According to balanced equation

mol of H2O formed = (6/2)* moles of H3PO4

= (6/2)*13.6

= 40.8 mol

Now use:

% yield = actual mol formed * 100 / theoretical number of mol

84.5 = actual mol formed * 100 / 40.8

actual mol formed = 34.5 moles

Answer: 34.5 moles