If 237.5 g of lead (II) oxide are produced in the following reaction, how many liters of oxygen are produced at STP
1 answer:
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<u>Answer:</u> The mole fraction of oxygen gas is 0.479
<u>Explanation:</u>
We are given:
Partial pressure of methane = 143 mmHg
Partial pressure of nitrogen gas = 469 mmHg
Partial pressure of oxygen gas = 563 mmHg
Total pressure = (143 + 469 + 563) = 1175 mmHg
To calculate the mole fraction of oxygen gas, we use the equation given by Raoult's law, which is:
where,
= partial pressure of oxygen gas = 563 mmHg
= total pressure = 1175 mmHg
= mole fraction of oxygen gas = ?
Putting values in above equation, we get:
Hence, the mole fraction of oxygen gas is 0.479