How many grams of nh3 can be produced from 4.08 mol of n2 and excess h2. express your answer numerically in grams?

Question

Jada Huber

Chemistry

5 June, 16:50

How many grams of nh3 can be produced from 4.08 mol of n2 and excess h2. express your answer numerically in grams?

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Answers (1)

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Yasmin Mullins · Answer 1

The reaction of N2 and H2 to generate NH3 is as follows:N2 + 3H2 ⇒2NH3

Since H2 is in excess amount in this reaction, the determine factor is N2. The ratio of N2 and product NH3 is 1:2. Therefore, 4.08 moles of N2 can generate 4.08*2 = 8.16 mole of NH3.

Grams of NH3 = moles of NH3 * molar mass of NH3 = 8.16 mole * 17 g/mole = 139 g.

139 grams of nh3 can be produced from 4.08 mol of n2 and excess h2.