Question

The combustion of pentane produced 6.58 g of carbon dioxide. What mass of pentane is required if the yield is 90 %? C5H12 (g) + 8O2 (g) 5CO2 (g) + 6H2O (l)

Answer 1

The mass of pentane required to produce 6.58 g of CO2 with a 90% yield can be calculated using stoichiometry and the molar masses of CO2 and pentane based on the balanced chemical equation for the combustion of pentane.

Step-by-step explanation:

The student is asking for the mass of pentane required to produce a certain amount of carbon dioxide, given a 90% yield of the reaction. To calculate this, we'll use the balanced chemical equation for the combustion of pentane:

C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(l)

The first step is to use the molar mass of CO2 to find out how many moles of CO2 correspond to 6.58 g. Then, using stoichiometry, we calculate the moles of pentane that would be needed for this amount of CO2. Finally, we adjust for the 90% yield by dividing the theoretical amount by 0.9 (yield percentage as a decimal). After calculating the moles of pentane, we convert this to grams using the molar mass of pentane.