how much methane or ammonia will be left when the reaction is finished?

Question

If 25 grams of methane, CH 4, and 30 g of ammonia, NH 3, are combined w/ excess oxygen, how much methane or ammonia will be left when the reaction is finished?

CH 4 + NH 3 + O 2 ® 2HCN + 6H 2O

Answer 1

The equation as given is NOT balanced. So, first we have to balance the equation ...

2CH₄ + 2NH₃ + 3O₂ ==> 2HCN + 6H₂O ... balanced equation

There is excess O2, but we still need to determine if CH4 or NH3 is limiting. We do this by dividing moles of CH4 by 2 and dividing moles of NH3 by 2 and see which is less.

For CH4: 25 g CH4 x 1 mol CH4 / 16 g = 1.56 moles CH4

For NH3: 30 g NH3 x 1 mol NH3 / 17 g = 1.76 moles NH3

CH4 is LIMITING

We now use the moles of the limiting reactant to find out how many moles of the excess reactant is used up. We then subtract that from the amount originally present to find how much of the excess reactant is left over.

moles NH3 used = 1.56 mols CH4 x 2 mol NH3 / 2 mols CH4 = 1. 56 moles NH3 used

moles NH3 left over = 1.76 mols - 1.56 mols = 0.20 mols NH3 left

mass of NH3 left over = 0.20 mols x 17 g / mol = 3.4 g NH3 left over

There will be no CH4 left over as that is the limiting reactant and will be all used up