Question

Methane (CH4), ammonia (NH3), and oxygen (O2) can react to form hydrogen cyanide (HCN) and water according to this equation:

CH4 + NH3 + O2 à 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.

Answer 1

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

-----> CH₄ = Whichever reactant produces the smaller amount of product is the limiting reagent. This is because the limiting reagent runs out before all of the other reactant is completely used. Because CH₄ produces the smaller amount of product, it is the limiting reagent.

-----> 13.5 grams HCN

Step-by-step explanation:

Part 1:

An equation is balanced when there is an equal amount of each reactant on both sides of the reaction. These amounts can be modified by adding coefficients in front of the molecules.

The unbalanced equation:

CH₄ + NH₃ + O₂ -----> HCN + H₂O

Reactants: 1 carbon, 7 hydrogen, 1 nitrogen, 2 oxygen

Products: 1 carbon, 3 hydrogen, 1 nitrogen, 1 oxygen

The balanced equation:

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

Reactants: 2 carbon, 14 hydrogen, 2 nitrogen, 6 oxygen

Products: 2 carbon, 14 hydrogen, 2 nitrogen, 6 oxygen

Part 2:

You can determine the limiting reactant by converting each reactant mass (besides O₂) to a product mass. Whichever reactant produces the smaller amount of product is the limiting reagent. This is because the limiting reagent runs out before all of the other reactant is completely used.

Let's convert to HCN because Part 3 also wants to know how much HCN is produced. To find this amount, you need to (1) convert grams reactant to moles reactant (via their molar masses), then (2) convert moles reactant to moles HCN (via the mole-to-mole ratio from equation coefficients), and then (3) convert moles HCN to grams HCN (via its molar mass).

Molar Mass (CH₄): 12.011 g/mol + 4(1.008 g/mol)

Molar Mass (CH₄): 16.043 g/mol

Molar Mass (NH₃): 14.009 g/mol + 3(1.008 g/mol)

Molar Mass (NH₃): 17.033 g/mol

Molar Mass (HCN): 1.008 g/mol + 12.011 g/mol + 14.009 g/mol

Molar Mass (HCN): 27.028 g/mol

8 g CH₄ 1 mole 2 moles HCN 27.028 g
-------------- x ---------------- x ---------------------- x ---------------- = 13.5 g HCN
16.034 g 2 moles CH₄ 1 mole

10 g NH₃ 1 mole 2 moles HCN 27.028 g
--------------- x ---------------- x ----------------------- x ---------------- = 15.9 g HCN
17.033 g 2 moles NH₃ 1 mole

Because CH₄ produces the smaller amount of product, it is the limiting reagent. Therefore, the actual amount of HCN produced is 13.5 g.