This would be a huge help

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

This same question has been asked and answered twice in the last few hours.

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

Excess reactant is O2. How much is left over? Subtract mols used up from mols originally present.

moles used up = 6.3 mols Mg x 1 mol O2 / 2 mols Mg = 3.15 mols used up

moles left over = 6.9 mols - 3.15 mols = 3.8 mols left over (2 sig. figs.)

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

To find limiting reactant just 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

To find mols of excess (H2O) reactant, find mols used up and subtract from mols originally present.

moles H2O used up = 3.4 mols Fe x 4 mols H2O / 3 mols Fe = 4.53 mols H2O used

moles H2O left over = 9.2 mols - 4.53 mols = 4.7 mols H2O left over (2 sig. figs.)