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?
responded above.
To determine the number of grams of O2 required to react completely with 3 moles of propane (C3H8), we will use the balanced equation for the combustion of propane:
C3H8 + 5O2 -> 3CO2 + 4H2O
First, we need to find the molar mass of propane (C3H8) and oxygen (O2). The molar mass of C3H8 is calculated by multiplying the atomic mass of carbon (C) by 3 and the atomic mass of hydrogen (H) by 8, and adding them together.
The atomic mass of carbon (C) is around 12.01 g/mol, and the atomic mass of hydrogen (H) is around 1.01 g/mol. Therefore, the molar mass of propane (C3H8) is:
3 * 12.01 g/mol + 8 * 1.01 g/mol = 44.11 g/mol
Next, we need to calculate the molar mass of oxygen (O2). The atomic mass of oxygen (O) is around 16.00 g/mol. Since we have 2 oxygen atoms in O2, the molar mass is:
2 * 16.00 g/mol = 32.00 g/mol
Now, we can determine the amount of O2 required to react with 3 moles of propane. According to the balanced equation, the mole ratio of C3H8 to O2 is 1:5. This means that for every 1 mole of C3H8, we need 5 moles of O2.
Since we have 3 moles of propane, we can multiply this by the mole ratio to find the moles of O2 required:
3 moles C3H8 * (5 moles O2 / 1 mole C3H8) = 15 moles O2
Finally, to find the mass of O2 required, we multiply the moles of O2 by its molar mass:
15 moles O2 * 32.00 g/mol = 480 g
Therefore, 480 grams of O2 are required to react completely with 3 moles of propane (C3H8).