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2 questions:

  1. 4 Fe(s) + 3O2(g) -->2Fe2O3(g)

In a certain reaction, 27.3 g of iron is reacted with 45.8 g of Oxygen


How many grams of the excess reactant remain after the reaction is complete

  1. N2 (g) + 3H2 (g) -->2 NH3 (g)

The equation above is the equation for the Haber process.


In a certain reaction, you start with 3.0 moles of nitrogen and 5.0 moles of hydrogen,. How many moles of ammonia will be produced in the reaction?

Verified Answer :

SapphireMoon
  1. So 27.3 g Fe is 0.4911 mol (27.3 / 55.58). 45.8 g O is 2.862 mol. Because of the stoichiometry, 3/4 as much oxygen as iron gets consumed. For the 0.4911 mol Fe consumed, 0.3683 mol O is consumed. 2.862-0.3863 = 2.4757 mol, or 39.6 g O is left.

    2. Hydrogen is the limiting reactant for these quantities. 2/3 as much ammonia is produced as hydrogen is consumed, so for 5.0 mol H2, that's 3.33 mol NH3.

    Make sense?

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