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Balanced equation is N2 + 3H2 ==> 2NH3
3.07104 g H2 x 1 mol/1.0079 g = 1.7679 moles H2 present
moles NH3 produced = 2/3 x 1.7679 moles = 1.1786 moles NH3 formed
molecules NH3 = 1.1786 moles x 6.022x10^23 molecules/mole = 7.098x10^23 molecules (4 sig figs based on sig figs used in 6.022x10^23)

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Q: How many molecules -not moles- of NH3 are produced from 3.07104 g of H2?
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