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Octane, C8H18, is one of the hydrocarbons in petrol. Upon combustion, octane produces C02rnand H20. What volume of oxygen measured at 101.3 kpa and 260c is required to burn 1.50 grnof octane? (mw = 114 g mol-l)
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Sure thing! Here's what I found:
The balanced chemical equation for the combustion of octane is:
C8H18 + 12.5O2 → 8CO2 + 9H2O
From this equation, we can see that 1 mole of octane requires 12.5 moles of oxygen for complete combustion.
To determine the volume of oxygen required, we can use the ideal gas law: PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (8.314 J/mol-K), and T is the temperature.
Given that the pressure is 101.3 kPa and the temperature is 260°C (533.15 K), we can rearrange the ideal gas law to solve for the volume:
V = nRT/P
First, we need to convert the mass of octane (1.50 g) to moles:
n = m/MW
n = 1.50 g / 114 g/mol
n = 0.0132 mol
Now we can calculate the volume of oxygen required:
V = nRT/P
V = (0.0132 mol)(8.314 J/mol-K)(533.15 K) / (101.3 kPa)
V = 0.584 L
Therefore, 0.584 L of oxygen at 101.3 kPa and 260°C is required to burn 1.50 g of octane.
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