Answer:
There will be produced 89.6 L of NH3
There are 5 moles of N2 given
Step-by-step explanation:
Step 1: Data given
Volume of N2 = 112 L
Volume of H2 = 134.4 L
Step 2: The balanced equation
N2 + 3 H2 → 2 NH3
Step 3: Calculate moles of H2
For 22.4 L we have 1 mol
Number of moles = 134.4 / 22.4 = 6 moles
Step 4: Calculate moles of N2
Number of moles = 112 / 22.4 = 5 moles
Step 5: Calculate limting reactant
For 1 mol of N2 we need 3 mol of H2 to produce 2 moles of NH3
The limiting reactant is H2. It will be completely be consumed ( 6 moles)
N2 is in excess. There will react 6 moles / 3 = 2 moles
There will remain 5 - 2 = 3 moles
Step 6: Calculate moles of NH3
For 1 mol of N2 we need 3 mol of H2 to produce 2 moles of NH3
For 6 moles of H2 we'll have 4 moles of NH3
Step 7: Calculate volume of NH3
1 mol = 22.4 L
4 moles = 89.6 L
There will be produced 89.6 L of NH3