How many molecules of NO can be produced by the reaction of 65.00g of NH3, with 150.0g of O2, according to the balanced equation? 4NH3(g) +5O2(g)=4NO(g)+6H2O(g) ?

Question

I know we have to find the limiting reactant and convert grams to moles but I am not sure how. Step by Step please and how do we get molecules of NO?

Answer 1

4NH₃(g) + 5O₂(g) ==> 4NO(g) + 6H₂O(g) ... balanced equation

You are correct that we must first find the limiting reactant.

For NH₃ we have:

65.00 g NH₃ x 1 mol NH₃/17 g x 4 mol NO/4 mol NH₃ x 30 g NO/mol = 114.7 g NO formed

For O₂ we have:

150.0 O₂ x 1 mol O₂/32 g x 4 mol NO/5 mol O₂ x 30 g NO/mol = 112.5 g NO formed <--LIMITING

Now that we have the limiting reactant (O2), and we've calculated the moles NO formed (112.5), we can convert this to molecules of NO.

112.5 g NO x 1mol/30 g = 3.75 mol x 6.02x10²³ molecules/mol = 2.258x10²⁴ molecules