Question

I need help.
Methane (CH4), ammonia (NH3), and oxygen (O2) can react to form hydrogen cyanide (HCN) and water according to this equation:
CH4 + NH3 + O2  HCN + H2O
You have 8 g of methane and 10 g of ammonia in excess oxygen.
How many grams of hydrogen cyanide will be formed? Show your work.

Answer 1

2CH4 + 2NH3 + 3O2 --> 2HCN + 6H2O

Answer 2

0 g of NH3 is 0.588 moles (10/17) and 8 g of CH4 is 0.5 moles (8/16).

Answer 3

You have excess oxygen, so you need to find the limiting reagent

Answer 4

0.588/2 = 0.294, and 0.5/2 = 0.25. Therefore, CH4 is your limiting reagent, and this is what you need to use to find the amount of HCN produced.

Answer 5

CH4 and HCN have the same coefficient in the equation

Answer 6

If you're using up all 0.5 moles of CH4, you are producing 0.5 moles of HCN. 0.5 moles of HCN is 13.5 g (0.5 * 27).

Answer 7

So the Balanced equation : 2CH4 + 2NH3 +302 --> 2HCN + 6H20

Answer 8

CH4=8/16=0.5 NH3=10/17=0.59 <- which is the number of moles

Answer 9

One mole of CH4 requires one mole of NH3.

Answer 10

There are 0.59 moles of NH3

Answer 11

0.5*27= 13.5 g.

Answer 12

Does that make sense?

Answer 13

yes, thank you very much!

Answer 14

No problem!