What volume of liquid water is produced from the combustion of 10.00 L of octane, C8H18? (density of octane=0.692 g/mL
To determine the volume of liquid water produced from the combustion of 10.00 L of octane (C8H18), we need to know the balanced chemical equation for the combustion reaction. In this case, the balanced equation is:
C8H18 + 12.5O2 → 8CO2 + 9H2O
From the balanced equation, we can see that for every 9 moles of water produced, we need 1 mole of octane.
First, we convert the volume of octane from liters to grams using the density of octane, which is 0.692 g/mL. Since 1 mL is equal to 1 cm³, we can convert liters to cm³ by multiplying by 1000:
10.00 L * 1000 cm³/L = 10,000 cm³
Next, we use the molecular weight of octane (114.23 g/mol) to convert the volume from cm³ to grams:
10,000 cm³ * 0.692 g/cm³ = 6,920 g
Now, we can calculate the number of moles of octane:
6,920 g / 114.23 g/mol = 60.597 mol
Since we know that 1 mole of octane produces 9 moles of water, we then calculate the number of moles of water produced:
60.597 mol * 9 mol H2O / 1 mol C8H18 = 545.373 mol H2O
Finally, we convert the moles of water produced to the volume of water in liters using the density of water, which is 1 g/mL:
545.373 mol H2O * 18.015 g/mol / 1 mL * 1 L/1000 mL = 9.817 L
Therefore, the combustion of 10.00 L of octane produces approximately 9.817 L of liquid water.