Answer:
168.64 grams of O2
Step-by-step explanation:
To burn 46.8 grams of propane (C3H8), we need to find the amount of oxygen that reacts with it according to the balanced chemical equation:
C3H8 + 5O2 -> 3CO2 + 4H2O
From the equation, 1 mole of propane (C3H8) reacts with 5 moles of oxygen to form 3 moles of carbon dioxide and 4 moles of water. The number of moles of propane can be calculated as follows:
46.8 g C3H8 / (44.1 g/mol) = 1.054 mol C3H8
Therefore, 1.054 mol of propane (C3H8) requires 5 x 1.054 = 5.27 moles of oxygen to burn completely. To find the mass of the oxygen in grams, we multiply the number of moles by the molar mass of oxygen (O2):
5.27 moles O2 * 32.0 g/mol = 168.64 g O2.
So, 168.64 grams of oxygen are required to burn 46.8 grams of propane (C3H8).