iron(III) oxide reacts with carbon monoxide gas to form solid iron metal and carbon dioxide gas:

Fe2O3 + 3 CO --> 2 Fe + 3 CO2

If you begin the reaction with 84.34 g of iron(III) oxide and 68.87 g of CO, which reactant will be in excess at the end of the reaction and how many grams will be remaining?

I work limiting reagent problems by solving TWO simple stoichiometry problems. Here is a solved example of a simple stoichiometry problem.

http://www.jiskha.com/science/chemistry/stoichiometry.html
Using the amount of Fe203 in the problem, solve for the amount of Fe that will be formed. Then use the amount of CO in the problem and solve for the amount of Fe. The SMALLER amount of Fe will be the correct amount and the reactant producing that value will be the limiting reagent. After identifying the limiting reagent, use that value to determine the amount of the non-limiting reagent will be used and subtract from the original amount to fine the grams left unreacted.

To determine which reactant is in excess and how many grams will be remaining, we need to calculate the number of moles of each reactant.

First, let's calculate the number of moles of iron(III) oxide (Fe2O3):
The molar mass of Fe2O3 = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol

Number of moles of Fe2O3 = mass / molar mass
Number of moles of Fe2O3 = 84.34 g / 159.70 g/mol = 0.5278 mol

Next, let's calculate the number of moles of carbon monoxide (CO):
The molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol

Number of moles of CO = mass / molar mass
Number of moles of CO = 68.87 g / 28.01 g/mol = 2.4575 mol

According to the balanced equation, the stoichiometric ratio between Fe2O3 and CO is 1:3. This means that 1 mole of Fe2O3 reacts with 3 moles of CO.

For the complete reaction of 0.5278 mol of Fe2O3, we would need (0.5278 mol) * (3 mol CO / 1 mol Fe2O3) = 1.5834 mol of CO.

Since we have 2.4575 mol of CO, which is greater than the amount needed (1.5834 mol), CO is in excess.

To find out how many grams of CO are remaining:
Remaining moles of CO = initial moles of CO - moles of CO used
Remaining moles of CO = 2.4575 mol - 1.5834 mol = 0.8741 mol

Mass of remaining CO = remaining moles of CO * molar mass of CO
Mass of remaining CO = 0.8741 mol * 28.01 g/mol = 24.440 g

Therefore, at the end of the reaction, carbon monoxide will be in excess with 24.440 grams remaining.

To determine which reactant is in excess and how many grams will remain, we need to compare the number of moles of each reactant and determine which one gets fully consumed.

First, let's calculate the number of moles for each reactant:

Number of moles of Fe2O3 = mass / molar mass
Fe2O3 molar mass = (2 x atomic mass of Fe) + (3 x atomic mass of O)
Fe2O3 molar mass = (2 x 55.845 g/mol) + (3 x 16.00 g/mol)
Fe2O3 molar mass = 159.69 g/mol

Number of moles of Fe2O3 = 84.34 g / 159.69 g/mol = 0.5283 mol

Number of moles of CO = mass / molar mass
CO molar mass = atomic mass of C + atomic mass of O
CO molar mass = 12.01 g/mol + 16.00 g/mol
CO molar mass = 28.01 g/mol

Number of moles of CO = 68.87 g / 28.01 g/mol = 2.4598 mol

Now, using the balanced chemical equation, we can determine the stoichiometric ratio between Fe2O3 and CO:

Fe2O3 + 3CO --> 2Fe + 3CO2

According to the equation, 1 mole of Fe2O3 reacts with 3 moles of CO.

Since we have 0.5283 moles of Fe2O3 and 2.4598 moles of CO, we can compare the ratios:

Fe2O3:CO = 0.5283 mol : 2.4598 mol

The ratio is approximately 1:4.66

Since the ratio is less than 1:3 (the stoichiometric ratio), Fe2O3 is the limiting reactant, and CO will be in excess.

To find how many grams of CO remain after the reaction, we need to calculate the amount of CO required to react with the available Fe2O3.

1 mole of Fe2O3 reacts with 3 moles of CO.
0.5283 moles of Fe2O3 will react with (3/1) x 0.5283 = 1.5849 moles of CO.

The remaining moles of CO = 2.4598 mol - 1.5849 mol = 0.8749 mol

To convert the remaining moles of CO to grams:
mass = moles x molar mass
mass = 0.8749 mol x 28.01 g/mol = 24.45 g

Therefore, after the reaction, 24.45 grams of CO will remain in excess.