what mass of oxygen would be required to completely burn 6.75 g of propane in the following reaction?
C3H8+5O2 => 3CO2+4H2O
* please help
To find the mass of oxygen required to completely burn 6.75 g of propane (C3H8), we will use stoichiometry.
In the balanced chemical equation, we see that the stoichiometric ratio between propane (C3H8) and oxygen (O2) is 1:5. This means that for every 1 mole of propane, we need 5 moles of oxygen to completely burn it.
To start, we need to find the number of moles of propane. We can use its molar mass to convert grams to moles.
The molar mass of propane (C3H8) can be calculated by adding up the atomic masses of its individual elements from the periodic table:
C = 12.01 g/mol (3 atoms)
H = 1.01 g/mol (8 atoms)
Molar mass of propane (C3H8) = (3 x 12.01 g/mol) + (8 x 1.01 g/mol) = 44.11 g/mol
Now, let's calculate the number of moles of propane:
moles of propane = mass of propane / molar mass of propane
moles of propane = 6.75 g / 44.11 g/mol
moles of propane ≈ 0.153 mol
Since the stoichiometric ratio between propane and oxygen is 1:5, we need five times the amount of moles of propane in moles of oxygen.
moles of oxygen = moles of propane x 5
moles of oxygen = 0.153 mol x 5
moles of oxygen = 0.765 mol
Finally, we need to convert moles of oxygen back to grams using its molar mass:
The molar mass of oxygen (O2) can be found in the periodic table:
O = 16.00 g/mol (2 atoms)
Molar mass of oxygen (O2) = (2 x 16.00 g/mol) = 32.00 g/mol
mass of oxygen = moles of oxygen x molar mass of oxygen
mass of oxygen = 0.765 mol x 32.00 g/mol
mass of oxygen ≈ 24.48 g
Therefore, approximately 24.48 grams of oxygen would be required to completely burn 6.75 grams of propane.
To determine the mass of oxygen required to completely burn 6.75 g of propane, you need to use the stoichiometry of the balanced chemical equation provided.
The balanced equation shows that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2) to produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O).
1 mole of any substance represents its molar mass in grams. The molar mass of propane (C3H8) can be calculated as follows:
(3 * atomic mass of carbon) + (8 * atomic mass of hydrogen)
Molar mass of carbon = 12.01 g/mol
Molar mass of hydrogen = 1.01 g/mol
Molar mass of propane = (3 * 12.01) + (8 * 1.01) g/mol
Molar mass of propane = 44.11 g/mol
Now, let's calculate the amount of oxygen required to react with 6.75 g of propane.
6.75 g of propane / (44.11 g/mol of propane) = 0.153 mol of propane
Since the ratio of propane to oxygen is 1:5 according to the balanced equation, we can multiply the moles of propane by 5 to get the moles of oxygen.
0.153 mol of propane * 5 = 0.765 mol of oxygen
Finally, to calculate the mass of oxygen required, we multiply the moles of oxygen by its molar mass.
0.765 mol of oxygen * (32.00 g/mol of oxygen) = 24.48 g
Therefore, approximately 24.48 grams of oxygen would be required to completely burn 6.75 g of propane.
Here is a worked example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html