Calculate the yield of NH3 in liters at STP, if you react 112 L of N2 with 134.4 L of H2, given the reaction: N2 + 3 H2 - -> 2 NH3 How many moles of N2 are given?

There will be produced 89.6 L of NH3

There are 5 moles of N2 given

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