Propane gas (c3h8) is combusted at a constant temperature and pressure. if 34.35l of propane react, what volume of oxygen is required? what volume of carbon dioxide is produced?

First, we must write the reaction equation for the combustion of propane:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Moreover, we note that under standard conditions, propane is a gas and under standard conditions, one mole of gas occupies 22.4 liters of volume. Using this, we can calculate the moles of propane present:
Moles propane = 34.35/22.4 = 1.53 moles

Next, the molar ratio, as visible from the reaction equation, between propane and oxygen is 1:5. We can use this to calculate the moles of oxygen required:
1.53 * 5 = 7.65 moles
Volume of oxygen = 7.65 * 22.4 = 171.36 liters

Next, using the same method, we calculate the moles of carbon dioxide then the volume of carbon dioxide:
Moles = 4.59
Volume = 102.82 liters

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml