Question

Consider the following balanced equation: 13O2(g) + 2C4H10(g) → 8CO2(g) + 10H2O(l)

If 2.90×102 moles of O2(g) and 27.1 moles of C4H10(g) are allowed to react, what is the theoretical yield of CO2(g)?

Answer 1

13 moles of O₂ reacts with 2 moles of C4H10

1 mole of O2 reacts with (2/13) moles of C4H10

2.90 X 10² moles of O2 reacts with (2 X 2.90 X 100/13) =44.61 moles of C4H10

But only 27.1 moles of C4H10 are available.So, C4H10 is the limiting reagent.So,we will calculate the yield with the given moles of C4H10.

2 moles of C4H10 gives 8 moles of CO2

1 mole of C4H10 gives (8/2) moles of CO2

27.1 moles of C4H10 gives (8 X 27.1/2) =108.4 moles of CO2

Now,

No of moles=given weight/molecular weight

108.4 mol= given weight/44 g/mol

Given weight=108.4 X 44= 4769.6 g