Question

How many moles of NH3 can be produced from 12.0 moles of H2 and excess N2? A. 6.0 moles B. 12.0 moles C. 18.0 moles D. 24.0 moles

Answer 1

From 8.0 moles of
`H_2` and excess
`N_2`, 8.0 moles of
`NH_3` can be produced. Option C is answer. Option C is answer.

Step-by-step explanation:

Write the balanced chemical equation.

`3H_2 + N_2` →
`2NH_3`

Identify the limiting reagent.

Since
`N_2`is in excess,
`H_2` is the limiting reagent.

Calculate the moles of NH3 produced using the balanced chemical equation and the moles of the limiting reagent.

moles of
`NH_3` = moles of
`H_2` * (2/3) = 12.0 * (2/3) = 8.0 moles

In conclusion, based on the given balanced chemical equation and the provided information about the reactants and excess ammonia, it is determined that from 12.0 moles of reactant A, 8.0 moles of product B (
`NH_3`) can be produced.

Option C is answer.

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Question with correct options

How many moles of NH3 can be produced from 12.0 moles of H2 and excess N2? A. 6.0 moles B. 12.0 moles C. 18.0 moles D. 24.0 moles

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