Respuesta :

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We can solve this problem when we use the conditions of a gas at standard temperature and pressure. It has been established that at STP where the temperature is 0 degrees Celsius and the pressure is 101.325 kPa, the volume of 1 mole of gas is 22.4 L. We will use this data for the calculations.

68.5 L ( 1 mol O2 / 22.4 L O2 ) = 3.06 mol O2

Explanation:

It is known that at STP, there are 22.4 L present in one mole of a substance.

Therefore, in 68.5 liters there will be 1 mol divided by 22.4 L times 68.5 L.

Mathematically,      [tex]68.5 L \times \frac{1 mol O_{2}}{22.4 L}[/tex]

                              = 3.05 mol

Hence, we can conclude that there are 3.05 moles present in 68.5 liters of oxygen gas at STP.

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