When iron (Fe) rusts (reacts with O2), the product is iron(III) oxide (Fe2O3).
If 25 g Fe is reacted with 25 g O2 in a sealed container, what are the masses of Fe, O2 and Fe2O3 in the container after the reaction is complete?
How is type of question calculated?
substances react by molar quantities ... so you need to convert to moles
find the moles of Fe and O2 ... by dividing by the molar masses
two Fe and three O react to form the product
there will be excess O2 remaining ... all the Fe will be reacted
find the moles of Fe2O3 formed , and the moles of O2 remaining
convert the moles to masses ... by multiplying by the molar masses
First, determine the reaction equation. You probably want
4Fe + 3O2 = 2Fe2O3
Fe is 55.8 g/mole
O is 16 g/mole
so,
25g Fe = 0.448 moles
25g O2 = 0.781 moles
In the reaction, Fe:O2 = 4:3
so, 0.448 moles Fe (0.224 moles of Fe2) requires 0.336 moles of O2
So there is not enough Fe to consume all the O2. There will be 0.445 moles of O2 left over.
So, the masses after the reaction are
Fe: 0
O2: 0.445*32 = 14.25g
Fe2O3: 0.224 * 159.6 = 35.75 g
Note that the total mass is still 50g
Here can you please tell me that can i use thus equation for rusting of iron Fe2O3 +3CO ----- 2Fe +3CO2
To calculate the masses of Fe, O2, and Fe2O3 after the reaction is complete, we need to use stoichiometry. Stoichiometry is a method used to determine the quantities of substances involved in a chemical reaction.
First, we need to write a balanced equation for the reaction:
4Fe + 3O2 -> 2Fe2O3
From the equation, we can see that 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3.
To calculate the moles of Fe and O2, we can use the formula: moles = mass / molar mass.
The molar mass of Fe is 55.85 g/mol, so the moles of Fe is calculated as follows:
Moles of Fe = 25 g Fe / 55.85 g/mol = 0.447 mol Fe
The molar mass of O2 is 32.00 g/mol, so the moles of O2 is calculated as follows:
Moles of O2 = 25 g O2 / 32.00 g/mol = 0.781 mol O2
Now, we need to determine which reactant is limiting. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product formed.
From the balanced equation, we can see that 4 moles of Fe react with 3 moles of O2. Therefore, the mole ratio between Fe and O2 is 4:3. Since we have 0.447 mol of Fe and 0.781 mol of O2, Fe is the limiting reactant because we have less Fe than O2 based on the mole ratio.
To determine the moles of Fe2O3 formed, we can use the mole ratio from the balanced equation (2 moles of Fe2O3: 4 moles of Fe):
Moles of Fe2O3 = (0.447 mol Fe / 4 mol Fe2O3) * 2 mol Fe2O3 = 0.2235 mol Fe2O3
Finally, to calculate the masses of Fe, O2, and Fe2O3, we can use their respective moles and molar masses:
Mass of Fe = Moles of Fe * Molar mass of Fe
Mass of O2 = Moles of O2 * Molar mass of O2
Mass of Fe2O3 = Moles of Fe2O3 * Molar mass of Fe2O3
Using the given molar masses:
Mass of Fe = 0.447 mol * 55.85 g/mol = 24.90 g
Mass of O2 = 0.781 mol * 32.00 g/mol = 25.00 g
Mass of Fe2O3 = 0.2235 mol * 159.69 g/mol = 35.68 g
Therefore, after the reaction is complete, the masses of Fe, O2, and Fe2O3 in the container would be approximately 24.90 g, 25.00 g, and 35.68 g, respectively.
To calculate the masses in this type of question, you need to:
1. Write a balanced equation for the reaction.
2. Determine the moles of reactants using their masses and molar masses.
3. Identify the limiting reactant by comparing the moles and the mole ratio from the balanced equation.
4. Calculate the moles of the product formed using the mole ratio.
5. Calculate the masses of reactants and the product using their respective moles and molar masses.