In the reaction 4 NH3 (g) + 5 O2 (g) -> 4 NO (g) + 6 H2O (l) The combination of 17.5 g NH3 and 26.2 g O2 produces what mass of NO?

Question

Can someone please help me with this answer I have been stuck on it for quite some time

Answer 1

First, write the balanced equation:

4NH₃(g) + 5O₂(g) ==> 4NO(g) + 6H₂O(l) ... balanced equation (fortunately this was given to you)

Next, find which reactant is limiting. One way to do this is to divided the moles of each reactant by its coefficient in the balanced equation and whichever value is less is the limiting reactant.

For NH₃: 17.5 g NH₃ x 1 mol NH₃ / 17 g = 1.029 moles NH₃ (÷4 -> 0.25)

For O₂: 26.2 g O₂ x 1 mol O₂ / 32 g = 0.8188 mols O₂ (÷5 -> 0.16). THIS IS LIMITING REACTANT

Finally, use the moles of the limiting reactant (O₂) to find the amount of product (NO) that can be formed.

0.8188 mols O₂ x 4 mols NO / 5 mols O₂ x 30 g NO / mol NO = 19.7 g NO formed