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. I have gotten as far as knowing that 10g of NH3 is 7.14 moles and 8 grams of CH4 is .5 moles of HCN but i do not know how to find the total grams of HCN

Question

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. I have gotten as far as knowing that 10g of NH3 is 7.14 moles and 8 grams of CH4 is .5 moles of HCN but i do not know how to find the total grams of HCN

anonymous · Answer

I NEED HELP FAST ONLY HAVE A 30 MIN LEFT

matt101 · Answer

2CH4 + 2NH3 + 3O2 --> 2HCN + 6H2O

10 g of NH3 is 0.588 moles (10/17) and 8 g of CH4 is 0.5 moles (8/16). You have excess oxygen, so you need to find the limiting reagent - do this by dividing each value by the coefficient in the equation, and the lowest one is your limiting reagent.

So, 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.

You can see that both CH4 and HCN have the same coefficient in the equation, so however many moles of CH4 is consumed is the amount of HCN produced. Therefore, if you're using up all 0.5 moles of CH4, you're producing 0.5 moles of HCN. 0.5 moles of HCN is 13.5 g (0.5 * 27).