Question

4. At 477 K and 645 mmHg, what volume of N2 will react completely with 29.2 L H2 to produce NH3? A. 24.4 L B. 12.2 L C. 7.40 L D. 9.7 L

Answer 1

T = 477 K

P = 645 mm Hg = 645/760 atm = 0.848 atm

V = ?

First, calculate amount of moles of H2

PV = nRT

n = PV/RT = 0.848*29.2 / (0.082) (477) = 0.633 mol of H2

From reaction

N2 + 3H2 <-> 2NH3 (this is balanced)

3 mol of H2 per each mol of N2 ... Since we have 0.66 mol of H2, expect 0.633/3 mol of N2, that is 0.211 mol of N2

But we need volume, not moles so:

PV= nRT

V = nRT/P = 0.211*0.082*477/(0.848) = 9.73 L

We need 9.73 liters of N2 to produc that amount