Consider the combustion of propane (C3H8) in the presence of oxygen: C3H8+5O2 => 3CO2+4H2O How many grams of O2 are required to react completely with 3 moles of propane?
Looks like it takes 5 moles of O2 to react with 1 mole of propane.
So you will need 15 moles of O2.
convert that to grams.
The balanced equation for the combustion of propane is:
C3H8 + 5O2 → 3CO2 + 4H2O
From the equation, we can see that 1 mole of propane reacts with 5 moles of oxygen. Therefore, to calculate the number of moles of oxygen required to react with 3 moles of propane, we multiply 3 moles of propane by the stoichiometric ratio of oxygen to propane:
3 moles C3H8 × (5 moles O2 / 1 mole C3H8) = 15 moles O2
Now, we can convert the moles of oxygen to grams by using the molar mass of oxygen (O):
15 moles O2 × (32 g/mol) = 480 grams O2
Therefore, 480 grams of O2 are required to react completely with 3 moles of propane.
To determine the amount of O2 required to react completely with 3 moles of propane, we need to use the balanced chemical equation for the combustion of propane.
The balanced equation is:
C3H8 + 5O2 => 3CO2 + 4H2O
According to the balanced equation, it takes 5 moles of O2 to react with 1 mole of propane (C3H8).
Since we have 3 moles of propane, we can set up a proportion to find the amount of O2 needed:
(5 moles O2) / (1 mole C3H8) = (x moles O2) / (3 moles C3H8)
Using the proportion, we can find the number of moles of O2 needed to react with 3 moles of propane:
x = (5 moles O2 * 3 moles C3H8) / (1 mole C3H8)
x = 15 moles O2 / 1 mole C3H8
x = 15 moles O2
So, 15 moles of O2 are required to react completely with 3 moles of propane.
To convert moles to grams, we need to use the molar mass of O2, which is approximately 32 g/mol.
To find the grams of O2 required, we can use the equation:
grams = moles * molar mass
grams O2 = 15 moles O2 * 32 g/mol = 480 grams O2
Therefore, 480 grams of O2 are required to react completely with 3 moles of propane.