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The molecular mass of NH3 is the sum of the Atomic Mass of nitrogen and three times the atomic mass of hydrogen, or 14.007 + 3(1.008) = 17.031. Therefore, the number of moles of NH3 in 14.0 grams is 14.007/17.031 = 0.822. Since each molecule of N2 supplies two nitrogen atoms and each molecule of NH3 needs only one nitrogen atom, the number of moles of N2 needed is half the number of moles of NH3 formed = 0.411.

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14y ago
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12y ago

This reaction?

2NaN3 -> 2Na + 3N2

15.00 grams N2 (1 mole N2/28.02 grams)(2 mole NaN3/3 moles N2)(65.02 grams/1 mole NaN3)

= 23.20 grams NaN3 needed

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9y ago

21.7 g of NaN3 are required to form 14.0 g of nitrogen gas.

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7y ago

14 g nitrogen is equal to 0,5 moles.

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8y ago

Approx 43.34 grams.

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11y ago

1 mole

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Q: How many moles of NH3 can be produced from 14 grams of N2?
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