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8,21L of propane gas ( C3H8) is burnt in sufficient oxygen at stp. How many liters of oxygen gas are consumed during combustion?

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Answer:

To determine the amount of oxygen gas consumed during the combustion of propane, we need to consider the balanced chemical equation for the combustion reaction:

C3H8 + 5O2 -> 3CO2 + 4H2O

From the balanced equation, we can see that for every one molecule of propane (C3H8) consumed, five molecules of oxygen (O2) are required.

Since we are given the volume of propane gas, which is 8.21L, and the reaction takes place at STP (Standard Temperature and Pressure), we can use the ideal gas law to calculate the volume of oxygen consumed.

At STP, one mole of any ideal gas occupies 22.4L. So first, we need to convert the volume of propane gas to moles using the molar volume:

(8.21L C3H8) / (22.4L/mol) = 0.3668 mol C3H8

Based on the balanced equation, the molar ratio between propane and oxygen is 1:5. Therefore, the moles of oxygen consumed will be:

0.3668 mol C3H8 * 5 mol O2 / 1 mol C3H8 = 1.834 mol O2

Finally, we can convert the moles of oxygen gas to liters at STP:

1.834 mol O2 * 22.4L/mol = 41.02L O2

Therefore, approximately 41.02 liters of oxygen gas are consumed during the combustion of 8.21 liters of propane gas at STP.

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