What is the maximum amount of NH3 gas formed by mixing of 14 g N2 and 6 g H2 gases is?

What is the maximum amount of NH3 gas formed by mixing of 14 g N2 and 6 g H2 gases is?
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N2(g) + 3H2(g) ==> 2NH3(g) is the balanced equation for thisreaction.14 g N2 x 1 mole N2/28 g = 0.5 moles N2 present6 g H2 x 1 mole H2/2 g = 3 moles O2 presentIn this situation, the amount of N2 is limiting the overallreaction. So, the maximum amount of NH3 that can be produced willbe limited by the amount of N2.0.5 moles N2 x 2 moles NH3/1 mole N2 = 1 moles NH3 is themaximum amount of NH3.
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