4. At 477 K and 645 mmHg, what volume of N2 will react completely with 29.2 L H2 to produce NH3?
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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
4. At 477 K and 645 mmHg, what volume of N2 will react completely with 29.2...
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