Question

How many moles are in 68.5 liters of oxygen gas at STP?

Answer 1

Step-by-step 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,
`68.5 L * (1 mol O_(2))/(22.4 L)`

= 3.05 mol

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

Answer 2

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