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

2. 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?

Question

Edward Kemp

Chemistry

26 April, 01:06

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

2. 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?

+1

Answers (1)

Know the Answer?

Kiersten Cooke · Answer 1

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?