4CH4 + 4NH3 + 6O2 ==> 4HCN + 12H2O
You will never balance by trial and error; here is a good site that will balance redox equations.
http://www.chemteam.info/Redox/Redox.html
mols NH3 = 10g/17 = about 0.6 but you need to do it more accurately.
mols CH4 8/16 = about 0.50
Convert mol NH3 and mols CH4 to mols HCN.
0.6mol NH3 x (4 mols HCN/4 mol NH3) = ?
0.0.5 mol CH4 x (4 mol HCN/4 mol CH4) = ?
You have two values for mols HCN and both can't be right. In limiting reagent problems the SMALLER number is ALWAYS the correct value.
Convert that value to g. g = mols x molar mass.
3. 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.
1 answer
1 answer