Answer:
The volume of carbon dioxide gas that is produced is 243L
Explanation:
This is the equation:
2C₇H₆(g) + 17O₂(g) → 14CO₂(g) + 6H₂O(g)
We have the mass of heptane that was burned, so let's determine the moles
Mass (g) / molar mass
We convert the mass to g → 0.140 kg . 1g / 1×10⁻³ kg = 140 g
140 g / 90 g/mol = 1.55 moles
Ratio is 2:14. 2 moles of heptane can produce 14 moles of CO₂
Therefore 1.55 moles, would produce (1.55 . 14) / 2 = 10.9 moles of CO₂
Finally we have to work with density if we want to know volume. Firstly we convert the moles of CO₂ to g.
10.9 moles of CO₂ . 44 g/mol = 479.6 g of CO₂
CO₂ density = CO₂ mass / CO₂ volume
CO₂ volume = CO₂ mass / CO₂ density
CO₂ volume = 479.6 g / 0,001976 g/mL → 242712.5 mL
Let's convert the volume to L
242712.5 mL . 1L/1000 mL = 242.71 L
Rounded to 3 significant digits → 243L
