Question

How many grams of ammonia would be produced from the reaction of 2.64 moles of hydrogen gas reacting with excess nitrogen?

Answer 1

29.92g

Step-by-step explanation:

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Answer 2

2.64 moles of H₂, reacting with excess N₂, produce around 29.98 grams of NH₃ according to the stoichiometry of the balanced chemical equation.

To determine the grams of ammonia (NH₃) produced from 2.64 moles of hydrogen gas (H₂) reacting with excess nitrogen (N₂), we use the stoichiometry of the balanced equation:

`N_2(g) + 3H_2(g)` →
`2NH_3(g)`

The balanced equation shows a 1:3 ratio between moles of hydrogen and moles of ammonia. Thus, for 2.64 moles of H₂, we can calculate the moles of NH₃ produced:

Moles of NH₃ = ((2.64 moles of H₂)/3)
`*` 2

Moles of NH₃
`\approx` 1.76

Next, we find the mass of NH₃ using its molar mass (approximately 17.03 g/mol):

Mass of NH₃ = Moles of NH₃
`*` Molar mass of NH₃

Mass of NH₃
`\approx` 1.76
`*` 17.03

Mass of NH₃
`\approx` 29.98 grams

Therefore, from 2.64 moles of H₂ reacting with excess N₂, approximately 29.98 grams of NH₃ would be produced.