anonymous
  • anonymous
How many grams of nitrogen are present in 46.34 g ammonium nitrate?
Chemistry
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
mmonium nitrate is NH4NO3 moleculear weight is 80 g / mole nitrogen weight per mole is 28 g / mole % N by weight in a mole = 100 * 28 /80 = 35% 46.34 g amonium nitrate is 46.34*0.35= 16.22 g N
anonymous
  • anonymous
@lily768 From where you get the 80

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