The balanced reaction for this will appear as:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
a. moles of NH3:
mol NH3 = 11 mol NH3
a. How many moles of NH3 can be produced from 16.5 mol of H 2 and excess N2?
b. How many grams of NH3 can be produced from 3.78 mol of N 2 and excess H2.
c. How many grams of H 2 are needed to produce 11.05 g of NH3 ?
d. How many molecules (not moles) of NH3 are produced from 2.92 × 10−4 g of H2?
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