Consider the reaction of methane with ammonia and oxygen.
2CH4 (g) + 2NH3 (g) + 3O2 (g) 2HCN (g) + 6H2O (l)
Determine the limiting reahelpctant in a mixture containing 123 g of CH4, 114 g of NH3, and 423 g of O2. Calculate the maximum mass (in grams) of hydrogen cyanide, HCN, that can be produced in the reaction.
The limiting reactant is:
NH3
CH4
O2
Amount of HCN formed =
g

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Answer :

Oseni Oseni · Answer 1 · 2022-08-25T12:16:19-04:00

The limiting reactant is [tex]NH_3[/tex]

The amount of HCN formed is 181.0593 grams

A limiting reactant determines the quantity of products formed in a reaction.

From the balanced equation of the reaction: [tex]2CH_4 (g) + 2NH_3 (g) + 3O_2 (g) --- > 2HCN (g) + 6H_2O (l)[/tex]

The mole ratio of the reactants is 2:2:3.

Mole of 123 g CH4 = 123/16 = 7.6875 moles

Mole of 114 g NH3 = 114/17 = 6.7059 moles

Mole of 423 g O2 = 423/32 = 13.2188 moles

Thus, NH3 appears to be the limiting reactant based on the number of moles and the mole ratio of the reactants.

The maximum amount of HCN that can be formed is determined by the amount of NH3.

Mole ratio of NH3 and HCN = 1:1

Equivalent mole of HCN = 6.7059 moles

Mass of 6.7059 moles HCN = 6.7059 x 27 = 181.0593 grams.

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