The combustion of propane may be described by the chemical equation
C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g)
How many grams of O2(g) are needed to completely burn 12.3 g C3H8(g)?
the coefficients in the reaction equation are moles
5 moles of O2 are needed to burn 1 mole of C3H8
find the number of moles in 12.3 g of C3H8
To determine the number of grams of O2(g) needed to completely burn 12.3 g of C3H8(g), we need to use stoichiometry, which is the quantitative relationship between the amounts of reactants and products in a chemical reaction.
The balanced chemical equation given is:
C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(g)
From the equation, we can see that for every 1 mole of C3H8, we need 5 moles of O2 to completely burn it.
To solve this problem, we can follow these steps:
Step 1: Convert the given mass of C3H8 to moles using its molar mass.
The molar mass of C3H8 is calculated by adding the atomic masses of carbon (C) and hydrogen (H):
Molar mass of C3H8 = (3*12.01 g/mol) + (8*1.01 g/mol) = 44.11 g/mol
Moles of C3H8 = Mass of C3H8 / Molar mass of C3H8
Moles of C3H8 = 12.3 g / 44.11 g/mol = 0.279 mol
Step 2: Use the stoichiometry of the balanced equation to find the moles of O2 needed.
From the balanced equation, we can see that for every 1 mole of C3H8, we need 5 moles of O2.
So the moles of O2 needed = Moles of C3H8 * (5 moles of O2 / 1 mole of C3H8)
Moles of O2 needed = 0.279 mol * 5 = 1.395 mol
Step 3: Convert moles of O2 to grams using its molar mass.
The molar mass of O2 is 32.00 g/mol.
Grams of O2 = Moles of O2 * Molar mass of O2
Grams of O2 = 1.395 mol * 32.00 g/mol = 44.64 g
Therefore, to completely burn 12.3 g of C3H8, we need 44.64 grams of O2.