Answer:
After complete reaction, 0.280 moles of ammonia are produced
Explanation:
Step 1: Data given
Number of moles N2 = 0.140 moles
Number of moles H2 = 0.434 moles
Step 2: The balanced equation
N2(g) + 3H2 (g) ⟶ 2NH3 (g)
Step 3: Calculate the limiting reactant
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
N2 is the limiting reactant. It will completely be consumed (0.140 moles).
H2 is in excess. There will react 3*0.140 = 0.420 moles
There will remain 0.434 - 0.420 = 0.014 moles
Step 4: Calculate moles NH3
For 0.140 moles N2 we'll have 2*0.140 = 0.280 moles NH3
After complete reaction, 0.280 moles of ammonia are produced
