If 2.25 g of nh 3 reacts with 3.38 g of o2 and produces 0.450 l of n 2 at 295 k and 1.00 atm , which reactant is limiting?

NH3 is the limiting reactant. O2 is in excess and there will remain 0.224 grams O2

Explanation:

Step 1: Data given

Mass of NH3 = 2.25 grams

Mass of O2 = 3.38 grams

Volume of N2 = 0.450 L

Temperature = 295 K

Pressure = 1.00 atm

Molar mass NH3 = 17.03 g/mol

Molar mass O2 = 32 g/mol

Molar mass of N2 = 28 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

4NH3 + 3O2 → 2N2 + 6H2O

Step 3: Calculate moles NH3

Moles NH3 = mass NH3 / molar mass NH3

Moles NH3 = 2.25 grams / 17.03 g/mol

Moles NH3 = 0.132 moles

Step 4: Calculate moles O2

Moles O2 = mass O2  / molar mass O2

Moles O2 = 3.38 grams / 32 g/mol

Moles O2 = 0.106 moles

Step 5: Calculate limiting reactant

For 4 moles NH3 we need 3 moles O2 to produce 2 moles N2 and 6 moles H2O

NH3 is the limiting reactant. It will completely be consumed (0.132 moles)

O2 is in excess. There will react 3/4 * 0.132 = 0.099 moles O2

There will remain 0.106 -0.099 = 0.007 moles O2

This is 0.007 moles * 32 g/mol = 0.224 grams

NH3 is the limiting reactant. O2 is in excess and there will remain 0.224 grams O2

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