calculate the grams of iron produced when 120g of iron (III) oxide are reacted with 84g carbon monoxide
Fe203+3CO--2Fe+3CO2
To calculate the grams of iron produced, we first need to determine the limiting reagent between iron (III) oxide (Fe2O3) and carbon monoxide (CO).
1. Calculate the number of moles of Fe2O3:
- Given mass of Fe2O3 = 120g
- Molar mass of Fe2O3 = 2(55.85 g/mol) + 3(16.00 g/mol) = 159.70 g/mol
- Moles of Fe2O3 = Mass of Fe2O3 / Molar mass of Fe2O3
= 120g / 159.70 g/mol
2. Calculate the number of moles of CO:
- Given mass of CO = 84g
- Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
- Moles of CO = Mass of CO / Molar mass of CO
= 84g / 28.01 g/mol
3. Calculate the mole ratio between Fe2O3 and CO:
From the balanced equation: Fe2O3 + 3CO -> 2Fe + 3CO2
The mole ratio between Fe2O3 and CO is 1:3.
4. Identify the limiting reagent:
The limiting reagent is the one that produces the least amount of product. It can be determined by comparing the mole ratios. Since the ratio is 1:3, for every mole of Fe2O3, we need 3 moles of CO. In this case, Fe2O3 is the limiting reagent because there are fewer moles of Fe2O3 (step 1) compared to what is required by the mole ratio.
5. Calculate the moles of Fe produced:
- Moles of Fe = Moles of limiting reagent (Fe2O3) x Mole ratio (from the balanced equation)
= (120g / 159.70 g/mol) x (2 moles Fe / 1 mole Fe2O3)
6. Convert moles of Fe to grams:
- Mass of Fe = Moles of Fe x Molar mass of Fe
= (120g / 159.70 g/mol) x (2 moles Fe / 1 mole Fe2O3) x 55.85 g/mol
7. Calculate the actual grams of Fe produced:
Since Fe2O3 is the limiting reagent, we can assume that all the Fe2O3 will react, and thus the grams of Fe produced will be equal to the calculated mass of Fe.
Therefore, the grams of iron produced when 120g of iron (III) oxide reacts with 84g carbon monoxide is calculated to be:
(120g / 159.70 g/mol) x (2 moles Fe / 1 mole Fe2O3) x 55.85 g/mol = 83.0 grams of Fe.
To calculate the grams of iron produced, we need to use stoichiometry to find the mole ratio between iron (III) oxide (Fe2O3) and iron (Fe).
First, we need to balance the equation:
Fe2O3 + 3CO → 2Fe + 3CO2
Next, we need to calculate the moles of each reactant:
Molar mass of Fe2O3 = (2 * atomic mass of Fe) + (3 * atomic mass of O)
= (2 * 55.845) + (3 * 16.00)
= 111.69 g/mol
Moles of Fe2O3 = mass of Fe2O3 / molar mass of Fe2O3
= 120 g / 111.69 g/mol
≈ 1.0746 mol
Molar mass of CO = atomic mass of C + atomic mass of O
= 12.01 + 16.00
= 28.01 g/mol
Moles of CO = mass of CO / molar mass of CO
= 84 g / 28.01 g/mol
≈ 3.0004 mol
Now, using the balanced equation, we can determine the mole ratio between Fe2O3 and Fe:
1 mol Fe2O3 : 2 mol Fe
Since the ratio is 1:2, we can conclude that for every 1.0746 moles of Fe2O3, we will produce 2 times that amount of Fe.
Moles of Fe = 2 * moles of Fe2O3
= 2 * 1.0746 mol
≈ 2.1492 mol
Finally, we calculate the grams of Fe produced:
Mass of Fe = moles of Fe * molar mass of Fe
= 2.1492 mol * 55.845 g/mol
≈ 120 g
Therefore, approximately 120 grams of iron will be produced when 120 grams of iron (III) oxide is reacted with 84 grams of carbon monoxide.