Question

A student places 142.9 moles C₅H₁₂ and 120.3 moles O₂ into a reaction chamber and ignited the mixture. How many moles CO2 were produced? Enter your answer to one place past the decimal and do not include units.

Answer 1

To calculate the moles of CO2 produced, we use the balanced combustion equation for C5H12, determine the limiting reagent, and apply stoichiometry. Given the amounts of C5H12 and O2, O2 is the limiting reagent, and 75.2 moles of CO2 are produced in the combustion reaction.

Step-by-step explanation:

To calculate the number of moles of CO2 produced when C5H12 is combusted, we first need to write down the balanced equation for the combustion:

C5H12 + 8 O2 → 5 CO2 + 6 H2O.

Based on the stoichiometry, each mole of C5H12 produces 5 moles of CO2. Now we must determine the limiting reagent.

Given:

142.9 moles C5H12

120.3 moles O2

From the balanced equation, we see that 1 mole of C5H12 reacts with 8 moles of O2, therefore, for 142.9 moles of C5H12 we would need 1143.2 moles of O2. Hence, O2 is the limiting reagent. From 120.3 moles of O2, we could calculate the moles of C5H12 that reacted as follows.

120.3 moles O2 × (1 mole C5H12/8 moles O2) = 15.0375 moles C5H12

Each mole of C5H12 produces 5 moles of CO2, so the moles of CO2 produced are:

15.0375 moles C5H12 × 5 = 75.2 moles CO2