Practice #2 questions

Question

1.      What is the greatest amount of magnesium oxide in moles that can be made with 6.3 mol Mg and 6.9 mol O₂? Which reactant is limiting? How many moles of the excess reactant will be left after the reaction?

2 Mg+ O2--> 2 MgO

2    How many moles of black iron oxide (Fe3O4) can be produced from 9.2 moles of water and 3.4 moles of iron? Which is the limiting reactant and how many moles of the excess reactant will be left after the reaction?

3 Fe + 4 H₂O ----> Fe₃O₄ + 4 H₂

Answer 1

2 Mg+ O₂ ==> 2 MgO ... balanced equation

Limiting reactant = Mg since it takes 2 mols Mg for each mol of O2, so not enough Mg.

Theoretical yield of MgO = 6.3 mols Mg x 2 mol MgO / 2 mols = 6.3 mols MgO

3Fe + 4H₂O ==> Fe₃O₄ + 4H₂ ... balanced equation

Easy way to find limiting reactant is to divide mols of each reactant by the corresponding coefficient in the balanced equation and see which is the least value.

For Fe we have 3.4 mols / 3 = 1.13

For H₂O we have 9.2 mols / 4 = 2.3

1.13 is less than 2.3 so Fe IS LIMITING.

To find mols of product, we use MOLES OF LIMITING REACTANT as follows:

3.4 mols Fe x 1 mol Fe₃O₄ / 3 mols Fe = 1.13 mols Fe₃O₄ can be produced