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

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.

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