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?
The carbon monoxide will be in excess at the end of the reaction, with 3.47 g remaining.
To determine which reactant is in excess, we need to compare the moles of each reactant and calculate the limiting reactant.
First, let's find the number of moles for each reactant:
Molar mass of Fe2O3 (Iron(III) oxide) = 2 * atomic mass of Fe + 3 * atomic mass of O
= 2 * 55.85 g/mol + 3 * 16.00 g/mol
= 159.69 g/mol
Number of moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
= 84.34 g / 159.69 g/mol
= 0.528 moles
Molar mass of CO (Carbon monoxide) = Atomic mass of C + Atomic mass of O
= 12.01 g/mol + 16.00 g/mol
= 28.01 g/mol
Number of moles of CO = Mass of CO / Molar mass of CO
= 68.87 g / 28.01 g/mol
= 2.459 moles
According to the balanced equation, the stoichiometric ratio between Fe2O3 and CO is 1:3. This means that 3 moles of CO are needed to react with 1 mole of Fe2O3. So, we need 3 * 0.528 moles of CO, which is equal to 1.584 moles.
Since we have 2.459 moles of CO and only need 1.584 moles, CO is in excess.
To find the mass of CO remaining, we need to subtract the mass of CO consumed from the initial mass of CO:
Mass of CO remaining = Initial mass of CO - Mass of CO consumed
= 68.87 g - (1.584 moles * 28.01 g/mol)
= 23.677 g
Therefore, at the end of the reaction, CO will be in excess, and there will be 23.677 grams of CO remaining.
To determine which reactant is in excess and how much of it will be remaining, we need to calculate the amount of each reactant needed for the reaction.
1. Calculate the molar mass of each compound:
- Iron(III) oxide (Fe2O3): 2(55.85 g/mol) + 3(16.00 g/mol) = 159.69 g/mol
- Carbon monoxide (CO): 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
2. Convert the given masses of iron(III) oxide and carbon monoxide to moles:
- Moles of Fe2O3: 84.34 g / 159.69 g/mol = 0.528 mol
- Moles of CO: 68.87 g / 28.01 g/mol = 2.457 mol
3. Determine the stoichiometry of the reaction:
According to the balanced equation, the molar ratio between Fe2O3 and CO is 1:3.
4. Calculate the amount of CO required to react with all the Fe2O3:
Moles of CO needed = 3 mol CO/mol Fe2O3 * 0.528 mol Fe2O3 = 1.584 mol CO.
5. Compare the amount of CO needed to the actual amount available:
Since we have 2.457 mol of CO, which is greater than 1.584 mol needed, CO is in excess.
6. Calculate the amount of excess CO remaining:
Excess mol of CO = 2.457 mol - 1.584 mol = 0.873 mol.
7. Convert the excess mol of CO back to grams:
Excess grams of CO = Excess mol of CO * molar mass of CO = 0.873 mol * 28.01 g/mol = 24.42 g.
Therefore, at the end of the reaction, carbon monoxide will be in excess with 24.42 grams remaining.