how many grams of CO are needed to react with an excess of Fe2O3 to produce 201.1 g Fe?
Fe2o3(s) + 3CO(g) -> 3CO2(g) + 2Fe(s)
To determine the amount of CO needed to react with an excess of Fe2O3 to produce 201.1 g Fe, we can use stoichiometry.
The balanced equation tells us that 1 mole of Fe2O3 reacts with 3 moles of CO to produce 2 moles of Fe.
To find the amount of CO needed, we need to follow these steps:
Step 1: Convert the given mass of Fe to moles:
Given:
Mass of Fe = 201.1 g
We know the molar mass of Fe is 55.85 g/mol.
Using the formula:
Moles = Mass / Molar mass,
we can calculate:
Moles of Fe = 201.1 g / 55.85 g/mol
Step 2: Use the mole ratio from the balanced equation to find the moles of CO:
From the balanced equation, we know that:
1 mole of Fe2O3 reacts with 3 moles of CO to produce 2 moles of Fe.
Since the ratio is 3 moles of CO: 2 moles of Fe,
we can set up the following proportion:
(3 moles of CO / 2 moles of Fe) = (x moles of CO / moles of Fe)
Solving for x, we get:
x = (3/2) * (moles of Fe)
Step 3: Convert the moles of CO to grams:
We need to use the formula:
Mass = Moles * Molar mass
Given:
Moles of CO = x
Molar mass of CO = 28.01 g/mol
Using the formula,
Mass of CO = Moles of CO * Molar mass of CO
Now, you can substitute the value of x and solve for the mass of CO.