N2 +3H2= 3NH3 What is the limiting reactant/ maximum number of moles of NH3 form

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Explanation / Answer

N2 MW = 28.0134 g/mole H2 MW = 2.0158 g/mole NH3 MW = 17.0304 g/mole N2(g) + 3H2(g) --> 2NH3(g) 1 g N2 = 1 g * mole/28.0134g = 0.0357 mole N2 3 g H2 = 3 g * mole/2.0158g = 1.46 mole H2 1 mole N2 needs 3 mole H2 so 0.0357 moles N2 need 3*0.0357 = 0.1071 mole H2 H2 is in excess so N2 is the limiting reagent now N2(g) --> 2NH3(g) 1 mole N2 produces 2 moles NH3 so 0.0357 moles NH3 produces 2*0.0357 = 0.0714 mole NH3 0.0714 mole NH3 = 0.0714 mole * 17.0304 g/mole = 1.2 g NH3

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