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How many moles of nh3 can be produced from 13.5 mol of h2 and excess n2?
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How many moles of nh3 can be produced from 13.5 mol of h2 and excess n2?

User Marurban
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Standard equation: N2(g)+3H2(g)==>2NH3(g), so through stoichiometry, (13.5 mol H2)(2 mol NH3/3 mol H2), yielding 9 moles of NH3
User Metaphore
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