what is the number of moles of Fe2O3 formed when 5.6liter of O2 reacts with 5.6g of Fe?
Balanced Reaction is:
4 Fe+3O2→2 Fe2O3
Given,
5.6liters of O2 participates in the reaction
1mole → 22.4 lt
x moles → 5.6lt
x=5.6/22.4 = 0.25 moles of O2
5.6g of Fe participates in the reaction
1mole → 56g
x moles → 5.6g
x=5.6/56 = 0.1 moles of Fe
In the reaction
4 Fe+3O2→2 Fe2O3
4 moles Fe reacts with 3 moles of O2 to give 2 moles of Fe2O3
As Fe has least mass it acts as a limiting reagent
So,
4 moles Fe gives 2 moles of Fe2O3= 2(0.1)/4
= 0.05 moles
To find the number of moles of Fe2O3 formed when 5.6 liters of O2 reacts with 5.6 grams of Fe, we will use the balanced chemical equation and the concept of stoichiometry.
The balanced chemical equation for the reaction between Fe and O2 to form Fe2O3 is:
4 Fe + 3 O2 → 2 Fe2O3
Step 1: Convert grams of Fe to moles of Fe
First, we need to calculate the number of moles of Fe using its molar mass. The molar mass of Fe is 55.85 g/mol.
5.6 g Fe * (1 mol Fe / 55.85 g Fe) = 0.1 mol Fe
Step 2: Convert liters of O2 to moles of O2
Next, we need to calculate the number of moles of O2 using the ideal gas law. The molar volume of any gas at STP (Standard Temperature and Pressure) is 22.4 liters.
5.6 L O2 * (1 mol O2 / 22.4 L O2) = 0.25 mol O2
Step 3: Determine the limiting reactant
To determine the limiting reactant, we compare the moles of Fe to the moles of O2. The balanced equation tells us that the ratio of Fe to O2 is 4:3. Therefore, we can say that every 4 moles of Fe require 3 moles of O2.
The ratio of moles of Fe to moles of O2 is:
0.1 mol Fe / 4 mol Fe = 0.025 mol O2 / 3 mol O2 = 0.0083
The smaller value is 0.0083, which means Fe is the limiting reactant.
Step 4: Calculate moles of Fe2O3
Since the balanced equation tells us that 4 moles of Fe react to form 2 moles of Fe2O3, we can use this ratio to determine the moles of Fe2O3 formed.
0.025 mol Fe * (2 mol Fe2O3 / 4 mol Fe) = 0.0125 mol Fe2O3
Therefore, 0.0125 moles of Fe2O3 are formed when 5.6 liters of O2 reacts with 5.6 grams of Fe.
To calculate the number of moles of Fe2O3 formed when 5.6 liters of O2 reacts with 5.6 grams of Fe, we need to use the balanced chemical equation for the reaction between Fe and O2. The balanced equation is:
4 Fe + 3 O2 -> 2 Fe2O3
First, we need to determine the number of moles of O2 using the ideal gas law equation:
PV = nRT
Where:
P = pressure (assuming it's at standard pressure, which is 1 atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (assuming it's at standard temperature, which is 273 K)
Given that the volume of O2 is 5.6 liters, we can substitute the values into the equation:
(1 atm)(5.6 L) = n(0.0821 L.atm/mol.K)(273 K)
Solving for n, we get:
n = (1 atm * 5.6 L) / (0.0821 L.atm/mol.K * 273 K)
n ≈ 0.248 moles
So, we have approximately 0.248 moles of O2.
Next, we need to determine the number of moles of Fe using the molar mass of Fe. The molar mass of Fe is 55.85 g/mol. Given that we have 5.6 grams of Fe, we can calculate the number of moles:
moles of Fe = (mass of Fe) / (molar mass of Fe)
moles of Fe = 5.6 g / 55.85 g/mol
moles of Fe ≈ 0.100 moles
Now that we have the number of moles of O2 and Fe, we can determine the limiting reactant.
In this case, Fe is the limiting reactant because it has the smallest number of moles compared to the stoichiometry of the reaction. According to the balanced equation, 4 moles of Fe react with 3 moles of O2 to produce 2 moles of Fe2O3.
Using the ratio from the balanced equation, we can calculate the number of moles of Fe2O3 formed:
moles of Fe2O3 = (moles of Fe) * (2 moles of Fe2O3 / 4 moles of Fe)
moles of Fe2O3 = 0.100 mol * (2/4)
moles of Fe2O3 ≈ 0.050 moles
Therefore, approximately 0.050 moles of Fe2O3 will be formed when 5.6 liters of O2 reacts with 5.6 grams of Fe.
assuming STP for the O2, you have 5.6/22.4 = 0.25 moles O2
5.6g = 5.6/55.85 = 0.10 moles Fe
If your reaction is
4Fe + 3O2 = 2Fe2O3
then the reaction consumes elements in the ratio
(moles O2)/(moles Fe) = 3/4
Your available reagents are in the ratio of O2/Fe = 0.25/0.10 = 5/2
So, there is excess O2, and you will wind up with 0.05 moles of Fe2O3, since it takes 2 Fe to produce 1 Fe2O3.