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.

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