Question

i>clicker: Calculate the volume of N2 necessary to react with  

9.0 L of H2 gas at 450 K and 5.00 atm.

a) 9.0 L

b) 3.0 L

c) 27.0 L

d) 1.0 L   

Answer 1

N₂ (g) + 3H₂ (g) -----------> 2NH₃ (g)

1 mole of nitrogen reacts with 3 moles of hydrogen gas

Given 9 litres of H₂ gas.

Temperature = 450 K

Pressure = 5 atm

Applying ideal gas equation for H₂

PV = nRT , here R is the gas constant = 0.0821 atm L / mol*K

=> 5*9 = n *0.0821*450

=> n = 1.22 moles of N₂

So , 3 moles of H₂ reacts with 1mole of N₂

1.22 moles of H₂ requires 1.22 /3 moles of N₂ = 0.406 moles

Again applying ideal gas equation for N₂ for the given conditions we get

PV = nRT

=> 5V = 0.406*0.0821*455

=> V = 3.03 litres = 3 litres approx. Ans