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Consider the following balanced reaction between hydrogen and nitrogen to form ammonia: 3H2(g) + N2(g)→2NH3(g) How...

Consider the following balanced reaction between hydrogen and nitrogen to form ammonia: 3H2(g) + N2(g)→2NH3(g) How many moles of NH3 can be produced from 18.0 mol of H2 and excess N2? Express the number of moles to three significant figures

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Answer #1

H2 is the limiting reagent here as N2 is used in excess.

Moles of H2 = 18.0 mol

From the balanced equation given in the problem, it is clear that 3 mol of H2 give 2 mol of NH3

Therefore, 18.0 mol of H2 give = (2 mol x 18.0 mol)/3 mol = 12.0 mol of NH3

Hence, the moles of NH3 that can be produced = 12.0 mol

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