• anonymous
ok, so sorry, buuuuut..... I need help with this desperately... 2. 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. Answer the following questions: • What is the balanced equation for this reaction? • Which reagent is limiting? Explain why. • How many grams of hydrogen cyanide will be formed? Show your work.
  • chestercat
I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this
and thousands of other questions

  • anonymous
1) Balanced equation : 2CH4 + 2NH3 +302 --> 2HCN + 6H20 2) CH4=8/16=0.5 NH3=10/17=0.59 (No. of moles) One mole of CH4 requires one mole of NH3. 0.5 moles of CH4 requires only 0.5 moles of NH3. But there are 0.59 moles of NH3 hence NH3 is in excess and CH4 is limiting reagent. 3) 0.5*27= 13.5 g.
  • anonymous
thank you xD

Looking for something else?

Not the answer you are looking for? Search for more explanations.