What mass of O2 would be needed to burn 11g of the propane gas?
How can I assist you with chemistry, James?
To determine the mass of O2 required to burn 11g of propane gas (C3H8), we need to use the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the balanced equation, we can see that 1 mole of propane (C3H8) reacts with 5 moles of oxygen gas (O2).
To calculate the mass of O2 needed, we will follow these steps:
Step 1: Convert the mass of propane (C3H8) to moles.
To do this, we need to know the molar mass of propane, which is the sum of the atomic masses of its elements:
(3 * atomic mass of Carbon) + (8 * atomic mass of Hydrogen) = (3 * 12.01) + (8 * 1.01) = 44.10 g/mol.
Moles of C3H8 = Mass (g) / Molar mass (g/mol)
Moles of C3H8 = 11 g / 44.10 g/mol
Moles of C3H8 = 0.25 mol
Step 2: Determine the moles of O2 needed.
According to the balanced equation, 1 mole of propane requires 5 moles of O2.
Moles of O2 = Moles of C3H8 * (5 moles O2 / 1 mole C3H8)
Moles of O2 = 0.25 mol * (5 mol O2 / 1 mol C3H8)
Moles of O2 = 1.25 mol
Step 3: Convert moles of O2 to mass.
To do this, we need to know the molar mass of oxygen gas (O2), which is 32.00 g/mol.
Mass of O2 = Moles of O2 * Molar mass of O2
Mass of O2 = 1.25 mol * 32.00 g/mol
Mass of O2 = 40 g
Therefore, to burn 11g of propane gas (C3H8), you would need approximately 40 grams of oxygen gas (O2).
C3H8 + 5O2 ---> 3CO2 + 4H2O
5 mols O2 for every mol of C3H8
a mol of O2 is 2*16 = 32 grams
a mol of C3H8 is 3*12+8 = 44 grams
so
11 grams of propane is 1/4 mol
so we need 5/4 mols of O2
(5/4)(32)