Question

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)

Answer 1

B) 22.4L of CO2 (g)

Step-by-step 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.