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4 votes
In the chemical equation:

C3H8 +502 3CO₂ + 4H₂O
How many grams of oxygen are required to burn 46.8 gram of propane (C3H8)

User Roomsg
by
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1 Answer

4 votes

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).

User Emir
by
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