Iron can be produced by the reduction of iron(III) Oxide by carbon monoxide:
Fe2O3 + 3CO -> 2FE +3CO2
a) what maximum mass of CO2 could be produced from the complete reduction of 14.7g of Fe2O3?
b) if 8.73g of CO2 gas was actually collected, what percentage yield of the reaction?
can you please go into detail i cant seem to understand it no matter how many people show me the work on how they do it.They all seem to do it differently and i just can't seem to get it.
the reaction equation is in moles
so you need to convert grams of iron oxide to moles
the molar mass is 159.7 (two iron plus 3 oxygen)
14.7 g / 159.7 g/mol = .092 moles
the reaction equation shows 3 moles of carbon dioxide for each mole of iron oxide
so .092 * 3 = .276 moles of CO₂
the molar mass of CO₂ is 44.0 g
a) .276 * 44.0 = 12.1 g
the percent yield is the actual amount divided by the theoretical amount
b) 8.73 g / 12.1 g = .721 = 72.1%
Sure, I'll be happy to explain the steps in detail.
a) To find the maximum mass of CO2 that could be produced from the complete reduction of 14.7g of Fe2O3, we need to use stoichiometry.
Step 1: Write down the balanced chemical equation.
Fe2O3 + 3CO -> 2Fe + 3CO2
Step 2: Determine the molar mass of Fe2O3 and CO2.
Fe2O3: Iron (Fe) has a molar mass of 55.85 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. Since there are 3 oxygen atoms in Fe2O3, the molar mass of Fe2O3 is (55.85 * 2) + (16.00 * 3) = 159.70 g/mol.
CO2: Carbon (C) has a molar mass of 12.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol. The molar mass of CO2 is 12.01 + (16.00 * 2) = 44.01 g/mol.
Step 3: Calculate the number of moles of Fe2O3.
To do this, divide the given mass (14.7g) by the molar mass of Fe2O3:
14.7g / 159.70 g/mol = 0.092 mol
Step 4: Determine the stoichiometric ratio between Fe2O3 and CO2.
From the balanced chemical equation, we see that for every 1 mole of Fe2O3, we produce 3 moles of CO2.
Step 5: Calculate the number of moles of CO2.
Multiply the number of moles of Fe2O3 (0.092 mol) by the stoichiometric ratio:
0.092 mol Fe2O3 * (3 mol CO2 / 1 mol Fe2O3) = 0.276 mol CO2
Step 6: Calculate the mass of CO2.
Multiply the number of moles of CO2 by its molar mass:
0.276 mol * 44.01 g/mol = 12.13 g
Therefore, the maximum mass of CO2 that could be produced from the complete reduction of 14.7g of Fe2O3 is 12.13 g.
b) To calculate the percentage yield of the reaction, we compare the actual yield (8.73g) to the maximum possible yield calculated in part a (12.13g).
Step 1: Calculate the percentage yield.
The percentage yield is given by the formula:
(Actual yield / Theoretical yield) * 100%
Step 2: Substitute the given values into the formula.
(8.73g / 12.13g) * 100% = 71.9%
Therefore, the percentage yield of the reaction is 71.9%.
Sure, I'll explain the process step by step in detail for you.
a) To find the maximum mass of CO2 produced from the complete reduction of 14.7g of Fe2O3, we first need to find the molar mass of Fe2O3 and CO2.
The molar mass of Fe2O3 can be calculated as follows:
Molar mass of Fe = 55.845 g/mol (from the periodic table)
Molar mass of O = 16.00 g/mol (from the periodic table)
Molar mass of Fe2O3 = 2 * Molar mass of Fe + 3 * Molar mass of O
= 2 * 55.845 g/mol + 3 * 16.00 g/mol
= 159.69 g/mol
Now, we can calculate the number of moles of Fe2O3 by using its molar mass and the given mass:
Number of moles = Mass / Molar mass
= 14.7 g / 159.69 g/mol
≈ 0.092 mol
According to the balanced chemical equation, 1 mole of Fe2O3 reacts with 3 moles of CO2. Therefore, the number of moles of CO2 produced can be calculated as follows:
Number of moles of CO2 = 3 * Number of moles of Fe2O3
= 3 * 0.092 mol
= 0.276 mol
Finally, we can find the maximum mass of CO2 produced by multiplying the number of moles by its molar mass:
Mass of CO2 = Number of moles of CO2 * Molar mass of CO2
= 0.276 mol * 44.01 g/mol (molar mass of CO2)
≈ 12.13 g
Therefore, the maximum mass of CO2 which could be produced from the complete reduction of 14.7g of Fe2O3 is approximately 12.13 g.
b) To calculate the percentage yield of the reaction, we need to compare the actual mass of CO2 collected to the maximum possible mass of CO2 calculated in part a.
Given that the actual mass of CO2 collected is 8.73g, we can calculate the percentage yield using the formula:
Percentage yield = (Actual mass / Maximum possible mass) * 100
Percentage yield = (8.73 g / 12.13 g) * 100
≈ 71.91%
Therefore, the percentage yield of the reaction is approximately 71.91%.
I hope this detailed explanation helps you understand the process of solving these types of problems.