Question

the reaction of 5.0g of pentane (C5H12) with a 5.0 g of oxygen gas produces 20.4 g of CO2 what is the percent yield of this reaction?

Answer 1

C5H12 + 8 O2 = 5 CO2 + 6 H2O

Molar mass of C5H12 = 72

Molar mass of O2 = 32

Molar mass of CO2 = 44

72g C5H12 ≡ 32x8 = 256g O2

∴ 5g C5H12 ≡ (5x256)/72 = 17.78 g O2,

but we have only 5g of O2

So, oxygen is the limiting reactant


Getting the theoretical yield

if 256 g O2 ≡ 5x44 g = 220 g CO2

∴ 5g O2 ≡ (5 x 220)/256 = 4.3 g


From these calculations 5 g of O2 can not produce 20.4 g of CO2, so, the percentage yield can not be calculated.

Answer 2

Answer:

Explanation:

C5H12 + 8 O2 = 5 CO2 + 6 H2O

Molar mass of C5H12 = 72

Molar mass of O2 = 32

Molar mass of CO2 = 44

72g C5H12 ≡ 32x8 = 256g O2

∴ 5g C5H12 ≡ (5x256)/72 = 17.78 g O2,

but we have only 5g of O2

So, oxygen is the limiting reactant

Getting the theoretical yield

if 256 g O2 ≡ 5x44 g = 220 g CO2

∴ 5g O2 ≡ (5 x 220)/256 = 4.3 g

From these calculations 5 g of O2 can not produce 20.4 g of CO2, so, the percentage yield can not be calculated.