Question

calculate the volume of NH3 gas formed when 0.25 dm³ of N2 is reacted fully with hydrogen?​

Answer 1

Answer:

Given,

vol. of N2 = 0.25 dm^3 = 0.25 Lit.

Chemical reaction :

N2 + 2H2 -------> 2NH3

we can infer that, One mole of N2 reacts with 2 moles of H2 to give 2 moles of NH2.

Here, vol. of N2 = 0.25 l

moles of N2 = given vol. /22.4 lit.

nN2 = 0.25/22.4 = 0.01 mole

Now,

no. of moles of H2 = 2 × nN2 = 0.01 × 2 = 0.02 moles

And, similarly, no. of moles of NH3 = 2 × nN2 = 0.02 moles

We know,

no. of moles of NH3 = vol. of NH3/Molar vol.

let vol. of NH3 be x,

so,

nNH3 = X/22.4

0.02 × 22.4 = x

or, x = 0.448 liters = 0.45 liters

Hence, vol. of NH3 formed = 0.45 Lit