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mol NH3 = 12 mol H2 x 2 mol NH3
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3 mol H2
mol NH3 = 8
mol NH3 = 12 mol N2 x 2 mol NH3
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1 mol N2
mol NH3 = 24
se agota el H2 Y en exceso el N2
mol = masa / Mm
Mm N2 = 28 g/mol
masa = mol x Mm
masa = 8 mol x 28 g/mol
masa de N2 = 224 g (respuesta d)