Question

Ammonia gas, NH3 (g) can be synthesized via the reaction:

2 NO (g) + 5 H₂ (g) -> 2 NH₃ (g) + 2 H₂O (g)

Suppose you start with a 45.8 g of NO and 12.4 g of H₂. Determine the limiting reactant and give the theoretical yield of ammonia, in grams. The molar mass of NO is 30.01 g/mol, and the molar mass of H₂ is 2.02 g/mol.

Answer 1

Answer – We are given, reaction –

2 NO (g) + 5 H₂ (g) ----> 2 NH₃ (g) + 2 H₂O (g)

Mass of NO = 45.8 g , mass of H₂ = 12.4 g

We need to calculate the moles of each reactant

Moles of NO = 45.8 g / 30.01 g.mol^-1 = 1.53 moles

Moles of H₂ = 12.4 g / 2.02 g.mol^-1 = 6.14 moles

Now limiting reactant –

Moles of NH₃ from NO

2 moles of NO = 2 mole of NH₃

So, 1.53 moles of NO = ?

= 1.53 moles of NH₃

Moles of NH₃ from H₂

5 moles of H₂ = 2 mole of NH₃

So, 6.14 moles of H₂ = ?

= 2.45 moles of NH₃

So moles of NH₃ got lowest from NO, so limiting reactant is NO.

Moles of NH₃ = 1.53 moles

So, theoretical yield of ammonia = 1.53 moles * 17.0307 g/mol

                                                       = 26.0 g