Final answer:
The complete combustion of 15.0 mL of heptane, given its density of 0.6838 g/mL, yields 27.6714 g of carbon dioxide. This is calculated by first finding the mass of heptane, converting it to moles, and then using the balanced combustion equation to determine the moles and finally the mass of CO2 produced.
Step-by-step explanation:
The question pertains to the combustion of heptane (C7H16) and requires finding the mass of carbon dioxide (CO2) produced when 15.0 mL of heptane burns completely. To solve this, we first need to find the mass of heptane using its given density (0.6838 g/mL), then write and balance the chemical equation for the combustion of heptane, which is C7H16 + 11O2 → 7CO2 + 8H2O. This indicates that every mole (114.23 g/mol for heptane) produces 7 moles of CO2. Therefore, once we have the mass of heptane, we can then use the mole ratio to calculate the moles and thus the mass of CO2 generated.
To calculate the mass of heptane: 15.0 mL × 0.6838 g/mL = 10.257 g.
Now using the molar mass of heptane (114.23 g/mol), we find the moles of heptane: 10.257 g ÷ 114.23 g/mol = 0.0898 mol. Applying the mole ratio from the balanced equation, we can calculate the moles of CO2 produced: 0.0898 mol × 7 mol CO2 = 0.6286 mol of CO2.
Finally, using the molar mass of CO2 (44.01 g/mol), we find the mass of CO2 produced: 0.6286 mol × 44.01 g/mol = 27.6714 g.
Therefore, 27.6714 g of CO2 is produced when 15.0 mL of heptane burns completely.