Question

Consider the following reaction. C₅H₁₂(g) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(g) If a container were to have 28 molecules of C₅H₁₂ and 28 molecules of O₂ initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion? (Use only whole numbers to solve.)

Answer 1

After the combustion reaction of C₅H₁₂ with O₂, considering the limiting reactant and using only whole numbers, we would have a total of 62 molecules (products and unreacted reactants) in the container.

Step-by-step explanation:

When analyzing the given reaction C₅H₁₂(g) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(g), we need to consider the stoichiometry to find out how many total molecules would be present after the reaction goes to completion. If we start with 28 molecules of C₅H₁₂ and 28 molecules of O₂, we see a ratio problem because C₅H₁₂ requires 8 molecules of O₂ for complete combustion. However, since there are only 28 O₂ molecules available, the reaction is limited by the oxygen available.

Since 28 O₂ molecules would fully react with 3.5 (= 28/8) molecules of C₅H₁₂, we will use only 3.5 molecules of C₅H₁₂. The products formed would be 3.5 * 5 = 17.5 molecules of CO₂ (round down to 17 for whole molecules) and 3.5 * 6 = 21 molecules of H₂O. But, considering whole numbers, again, you would need to adjust the C₅H₁₂ molecules to 3 (since it needs to be whole), which results in 15 CO₂ and 18 H₂O molecules being formed. Adding the leftover C₅H₁₂ (28 - 3 = 25) and unreacted O₂ (28 - 3 * 8 = 4), the total number of molecules would be 15 CO₂ + 18 H₂O + 25 C₅H₁₂ + 4 O₂ = 62 molecules after the reaction.