Answer:
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.
