Question

"For the reaction

C₇H₁₆(l) + 11O₂(g) → 7CO₂(g) + 8H₂O(l)
If 27.3g of C₇H₁₆ is combusted, what amount of Carbon Dioxide will be produced?"

Answer 1

To find the amount of carbon dioxide produced from the combustion of 27.3g of C7H16, we calculate the moles of n-heptane, use the stoichiometric ratio to determine moles of CO2, and then convert this to grams, yielding approximately 84.00 grams of CO2.

Step-by-step explanation:

The student is asking about the amount of carbon dioxide produced when 27.3 grams of C7H16 (n-heptane) is combusted. To answer this, we need to apply stoichiometry using the balanced chemical equation for the combustion of n-heptane: C7H16(l) + 11O2(g) → 7CO2(g) + 8H2O(l). First, we find the molar mass of C7H16 (100.2 g/mol) and CO2 (44.01 g/mol). Then we convert the mass of n-heptane to moles, and use the mole ratio from the balanced equation to calculate the moles of carbon dioxide produced. Finally, we convert moles of CO2 back to grams.

Steps as follows:

1. Calculate moles of C7H16: (27.3 g) / (100.2 g/mol) = 0.2727 moles C7H16
2. Use the mole ratio to find moles of CO2: 0.2727 moles C7H16 x (7 moles CO2 / 1 mole C7H16) = 1.909 moles CO2
3. Convert moles of CO2 to grams: 1.909 moles CO2 x (44.01 g/mol) = 84.0049 g CO2

Therefore, the mass of carbon dioxide produced when 27.3g of C7H16 is combusted is approximately 84.00 grams.