Question

Nitric Oxide is made from the oxidation of ammonia. What mass of nitric oxide can be made from the reaction of 8.00 g NH₃ with 17.0 g O₂?

4 NH₃(g) + 5 O₂(g) --> 4 NO(g) + 6 H₂O(g)

Answer 1

Answer – Given, mass of NH₃(g) = 8.00 g , mass of O₂(g) = 17.0 g

First we need to calculate the moles of each reactant

Moles of NH₃(g) = 8.00 g / 17.0307 g.mol^-1

                            = 0.470 moles

Moles of O₂(g) = 17.0 g / 32.0 g.mol^-1

               = 0.531 moles of O₂

The next step is determining the limiting reactant and moles of product

Moles of NO(g) from NH₃(g)

From the reaction –

4 moles of NH₃(g) = 4 moles of NO(g)

So, 0.470 moles of NH₃(g) = ?

= 0.470 moles of NO(g)

Moles of NO(g) from O₂(g)

From the reaction –

5 moles of O₂(g)= 4 moles of NO(g)

So, 0.531 moles of O₂(g)= ?

= 0.425 moles of NO(g)

So moles of No(g) is lowest from the O₂(g), so limiting reactant is O₂(g) and

Moles of NO(g) = 0.425 moles

Mass of NO(g) = 0.425 moles * 30.006 g/mol

                         = 12.75 g of NO(g)

So, 12.75 g of mass of nitric oxide can be made from the reaction of 8.00 g NH₃ with 17.0 g O₂.