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. A student has 8 g of methane and 10 g of ammonia in excess oxygen.

a. What is the balanced equation for this reaction?
b. Which reagent is limiting? Explain why.
c. How many grams of hydrogen cyanide will be formed?

Answer 1

Answer: a.
`2CH_4+2NH_3+3O_2\rightarrow 2HCN+6H_2O(g)`

b.
`CH_4` is the limiting.

c. 13.5 g of hydrogen cyanide will be formed

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of methane}=(8g)/(16g/mol)=0.50moles`

`\text{Moles of ammonia}=(10g)/(17g/mol)=0.59moles`

The balanced reaction will be :

`2CH_4+2NH_3+3O_2\rightarrow 2HCN+6H_2O(g)`

According to stoichiometry :

2 moles of
`CH_4` require = 2 moles of
`NH_3`

Thus 0.5 moles of
`CH_4` will require=
`(2)/(2)* 0.5=0.5moles` of
`NH_3`

Thus
`CH_4` is the limiting reagent as it limits the formation of product and
`NH_3` is the excess reagent as it is present in more amount than required.

As 2 moles of
`CH_4` give = 2 moles of
`HCN`

Thus 0.5 moles of
`CH_4` give =
`(2)/(2)* 0.5=0.5moles` of
`HCN`

Mass of
`HCN=moles* {\text {Molar mass}}=0.5moles* 27g/mol=13.5g`

Thus 13.5 g of
`HCN` will be produced from the given masses of both reactants.