Question

The combustion of pentane (C5H12)forms carbon dioxide (CO2) and water (H2O). The balanced chemical reaction is

C5H12(l)+8O2(g)⟶5CO2(g)+6H2O(l)C5H12(l)+8O2(g)⟶5CO2(g)+6H2O(l)

How many moles of carbon dioxide gas are produced when a 9.36 mol9.36 mol sample of pentane undergoes combustion with a 67.4%67.4% yield.

a 1.26 mol

b. 46.8 mol

c. 6.31 mol

d. 31.5 mol

Answer 1

When 9.36 moles of pentane undergo combustion with a 67.4% yield, 31.5 moles of CO2 are produced.

Step-by-step explanation:

To calculate the moles of carbon dioxide produced in the combustion of pentane (C5H12), we first look at the balanced equation C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l). From this, it is clear that 1 mole of pentane yields 5 moles of CO2 when completely combusted. If we start with 9.36 moles of pentane, we would expect 9.36 × 5 = 46.8 moles of CO2 in a 100% yield.

However, we are given a 67.4% yield, so the actual moles of CO2 produced are 67.4% of the theoretical yield. So we calculate 46.8 moles of CO2 × 67.4% = 31.5 moles of CO2. Thus, the correct answer is (d) 31.5 moles of CO2 produced.