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

Chemistry

Caylee Pace

4 September, 01:03

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?

+5

Answers (1)

Know the Answer?

Leila Stanton · 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?