--Statement--

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane:

2NH

_{3}(g) + 3O

_{2}(g) + 2CH

_{4}(g) --> 2HCN(g) + 6H

_{2}O(g)

--Question--

If 5.00 x 103 kg each of NH

_{3}, O

_{2}, and CH

_{4} are reacted, what mass of HCN and of H

_{2}O will be produced, assuming 100% yield?

--My process--

I assumed that one of the three reactants had to be limiting. I first found the number of moles of each, which came out to be 293530.586 mol NH

_{3}, 156250 mol O

_{2}, and 311681.835 mol CH

_{4}. Then I set up two ratios of available moles:

293530.586 mol NH_{3} =

1.8785957 mol NH_{3} 156250 mol O2 1 mol O2

311681.835 mol CH_{4} =

1.9947637 mol CH_{4} 156250 mol O2 1 mol O2

I compared these ratios to the ratios of required moles (derived from the balanced equation):

.6666667 mol NH_{3} .6666667 mol CH_{4} 1 mol O2 1 mol O2

Because the available moles of NH

_{3 } and of CH

_{4} were both in excess, I assumed that the O

_{2} would be consumed first, and would therefore be the limiting reactant. Next, I used the moles of the limiting reactant (O

_{2}) to find the masses of the products:

156250 mol O2x

2 mol HCN x

27.028 g HCN = 2.82 x 103 kg HCN

3 mol O2 1 mol HCN

156250 mol O2 x

6 mol H2O x

18.015 g H2O = 5.63 x 103 kg H2O

3 mol O2 1 mol H2O

I thought I solved the problem correctly, but my solution cannot be valid because it does not follow the Law of Conservation of Mass; the mass of the reactants (5.00 x 103 kg * 3 = 1.5 x 104 kg) does not equal the mass of the products (8.45 x 103 kg). I don’t know whether my approach to the problem is completely off. For example, maybe I should have set up a third mol ratio comparing NH3 and CH4. Also, I don’t know how the 100% yield will influence the problem. It means that the theoretical yield and actual yield are equal, but does that mean all the reactants will be consumed (i.e., there will be no limiting reactant?). I am just very confused and need some help. Thanks for your patience!