If 5.12 g of oxygen O2 gas occupies a volume of 6.21L at a certain temperature and pressure, how many grams of oxygen gas will occupy 30.3 L under the same conditions ?

Answer : The mass of occupy 30.3 L under the same conditions will be, 24.9 grams.

Explanation :

First we have to calculate the moles of

Now we have to calculate the moles of in 30.3 L by using Avogadro's law.

Avogadro's law : It is defined as the volume of gas is directly proportional to the number of moles of gas at constant pressure and temperature.

or,

where,

= initial volume of gas = 6.21 L

= final volume of gas = 30.3 L

= initial moles of gas = 0.16 mol

= final temperature of gas = ?

Now put all the given values in the above equation, we get:

Now we have to calculate the mass of

Molar mass of = 32 g/mol

Therefore, the mass of occupy 30.3 L under the same conditions will be, 24.9 grams.

14,16 18 reacts with group