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?
Well, it seems like we have a classic chemistry battle going on here! In the left corner, we have 84.34 g of iron(III) oxide. And in the right corner, we have 68.87 g of carbon monoxide gas. Let's see who will come out as the winner!
To compare the two, we need to find the molar masses of iron(III) oxide and carbon monoxide. After crunching some numbers, we find that the molar mass of Fe2O3 is approximately 159.7 g/mol, and the molar mass of CO is about 28 g/mol.
Now, let's determine how many moles of each we have. To do that, we divide the masses by their respective molar masses. For iron(III) oxide, we divide 84.34 g by 159.7 g/mol, which gives us about 0.527 mol. For carbon monoxide, we divide 68.87 g by 28 g/mol, obtaining approximately 2.461 mol.
According to the balanced equation, the stoichiometric ratio between iron(III) oxide and carbon monoxide is 1:3. That means for every 1 mol of Fe2O3, we need 3 mol of CO. In this case, we have slightly more moles of carbon monoxide (2.461 mol) compared to iron(III) oxide (0.527 mol).
Therefore, iron(III) oxide will be the reactant in excess, and some of it will be remaining after the reaction. To find out how much iron(III) oxide is left, we need to determine how many moles are required for the reaction to consume all the carbon monoxide.
Since the stoichiometric ratio is 1:3, for every 3 mol of CO, we need 1 mol of Fe2O3. Hence, if we have 2.461 mol of CO, we would need approximately 0.820 mol of Fe2O3. However, we only have 0.527 mol of Fe2O3, which is less than what is required. So, we can conclude that all the carbon monoxide will be consumed, but some iron(III) oxide will be left unreacted.
To find the remaining mass of iron(III) oxide, we multiply the moles by the molar mass. So, 0.527 mol multiplied by 159.7 g/mol gives us approximately 84.13 g of iron(III) oxide remaining.
Therefore, the reactant in excess is carbon monoxide, and approximately 84.13 g of iron(III) oxide will be left after the reaction, looking all sad and unreacted.
To determine which reactant is in excess at the end of the reaction, we need to compare the amount of each reactant used to the stoichiometric ratio of the reaction.
Let's start by calculating the number of moles for each reactant:
Molar mass of Fe2O3 (Iron(III) oxide) = 55.845 g/mol (2 x 55.845 + 3 x 16.00)
Molar mass of CO (Carbon monoxide) = 28.01 g/mol (12.01 + 16.00)
Number of moles of Fe2O3 = 84.34 g / (55.845 g/mol) = 1.509 mol
Number of moles of CO = 68.87 g / (28.01 g/mol) = 2.458 mol
Now, let's use the stoichiometric ratio to determine how many moles of CO are required to react with the given moles of Fe2O3:
Based on the balanced equation: 1 mole Fe2O3 reacts with 3 moles CO.
Therefore, the stoichiometric ratio of Fe2O3 to CO is 1:3.
From the above ratio, we can see that for every 1 mole of Fe2O3, we need 3 moles of CO.
Since we have 1.509 moles of Fe2O3 and 2.458 moles of CO, the amount of CO required (in moles) would be
1.509 mol Fe2O3 x (3 mol CO / 1 mol Fe2O3) = 4.527 mol CO
Comparing this to the 2.458 moles of CO we have, we can conclude that CO is the limiting reactant and Fe2O3 is in excess.
To find the amount of excess Fe2O3 remaining, we need to calculate the moles of CO that actually react:
2.458 mol CO - 1.509 mol Fe2O3 x (3 mol CO / 1 mol Fe2O3) = 1.429 mol CO remaining after the reaction.
Now, let's convert this amount back to grams:
Mass of excess Fe2O3 remaining = 1.429 mol CO x (28.01 g/mol) = 40.0 g
Therefore, at the end of the reaction, carbon monoxide (CO) will be in excess, and there will be 40.0 grams of iron(III) oxide (Fe2O3) remaining.
To determine which reactant is in excess at the end of the reaction, we need to calculate the amount of each reactant required to react completely.
First, we'll start by finding the molar mass of Fe2O3 (iron(III) oxide) and CO (carbon monoxide):
Molar mass of Fe2O3:
Fe = 55.845 g/mol
O = 16.00 g/mol (3 oxygen atoms, 16.00 g/mol each)
Molar mass of Fe2O3 = (2 * Fe) + (3 * O) = (2 * 55.845 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol
Next, we can calculate the number of moles for each reactant:
Number of moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
= 84.34 g / 159.69 g/mol
= 0.5282 mol
Number of moles of CO = Mass of CO / Molar mass of CO
= 68.87 g / 28.01 g/mol
= 2.4603 mol
Now, we look at the balanced chemical equation:
Fe2O3 + 3 CO --> 2 Fe + 3 CO2
According to the stoichiometry of the balanced equation, Fe2O3 and CO react in a 1:3 ratio. This means that for every 1 mole of Fe2O3, we need 3 moles of CO.
Here, we have 0.5282 moles of Fe2O3 and 2.4603 moles of CO. Since the molar ratio is 1:3, the limiting reactant will be Fe2O3 since we have less of it compared to the stoichiometric requirement.
To determine the amount of excess reactant remaining, we need to find out how much of the limiting reactant is consumed.
From the balanced equation, we know that for every 1 mole of Fe2O3, we'll have 3 moles of CO consumed. Therefore, the number of moles of CO consumed will be 3 * 0.5282 mol = 1.5846 mol.
Now, we can calculate the mass of CO consumed:
Mass of CO consumed = Moles of CO consumed * Molar mass of CO
= 1.5846 mol * 28.01 g/mol
= 44.346 g
Since we started with 68.87 g of CO, and 44.346 g was consumed, we can find the mass of CO remaining:
Mass of CO remaining = Initial mass of CO - Mass of CO consumed
= 68.87 g - 44.346 g
= 24.524 g
Therefore, at the end of the reaction, iron(III) oxide will be completely consumed, and 24.524 g of CO will be remaining as the excess reactant.