3) At STP, a 22.4-liter sample of NH3 (g) contains the same number of molecules as A) 11.2 L of H2 (g) B) 22.4 L of CO2 (g) C) 33.6 L of CH4 (g) D) 44.8 L of O2 (g)

B) 22.4L of CO2 (g)

Explanation:

Given data

At STP,

the volume of each gas is 22.414 dm³ (1 dm³ = 1 L),

one mole of a gas = 22.414 dm³ = 22.414 litre,

according to Avagadro's number:

one mole of a gas = 6.02 x 10^23 molecules.

Solution:

One litre of NH3 = (6.02x 10^23) / (22.414)

=2.68 x10^22 molecules of NH3.

22.4 litre of NH3 = 22.4 x 2.68 x10^22

= 6.01x 10^23 molecules of NH3.

A)

One litre of H2 = (6.02x 10^23) / (22.414)

=2.68 x10^22 molecules of H2.

11.2 litre of H2 = 11.2x 2.68 x10^22

= 3.01x 10^23 molecules of H2.

Similarly,

B)

One litre of CO2 = (6.02x 10^23) / (22.414)

=2.68 x10^22 molecules of CO2.

22.4 litre of CO2 = 22.4 x 2.68 x10^22

= 6.01x 10^23 molecules of CO2.

From here we see that option B is the correct one.