What volume of oxygen is required to completely combust 80 of butane
80 what?
80 grams
80 bushels
80 gallons
80 baseball fields
To determine the volume of oxygen required to completely combust butane, you need to know the balanced chemical equation for the combustion of butane, as well as the molar ratio between butane and oxygen.
The balanced chemical equation for the combustion of butane (C4H10) is:
2 C4H10 + 13 O2 -> 8 CO2 + 10 H2O
From the balanced equation, we can see that 2 moles of butane react with 13 moles of oxygen to produce 8 moles of carbon dioxide and 10 moles of water.
First, convert the given mass of butane (80 g) to moles by using its molar mass:
Molar mass of C4H10 = (4 x 12.01 g/mol) + (10 x 1.01 g/mol) = 58.12 g/mol
Moles of butane = 80 g / 58.12 g/mol = 1.3773 mol
Next, using the mole ratio from the balanced equation, determine the moles of oxygen needed:
Moles of oxygen = 1.3773 mol butane x (13 mol O2 / 2 mol butane) = 8.954 mol O2
Finally, convert the moles of oxygen to the volume at a given temperature and pressure, typically measured in liters (L). The conversion factor is the ideal gas law:
1 mol of any gas = 22.4 L (at standard temperature and pressure)
Volume of oxygen = 8.954 mol O2 x (22.4 L / 1 mol) = 200.43 L
Therefore, approximately 200.43 liters of oxygen are required to completely combust 80 grams of butane.