Question

Iron metal can be obtained from iron ore, iron (III) oxide. When the iron (III) oxide is reacted with carbon monoxide gas it forms iron metal and carbon dioxide gas. If 75.0 grams of iron (III) oxide are reacted with 50.0 g of carbon monoxide gas, what is
the theoretical yield of iron metal that can be produced?
Equation: ?
Limiting Reactant: ?
Excess: ?
Grams of target product produced:

Answer 1

The theoretical yield of iron metal is 42.3 g

The equation for the reaction is:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Since, all of the carbon monoxide was consumed during the reaction, then carbon monoxide is the limiting reactant

Thus, iron (III) oxide is the excess reactant

Stoichiometry

From the question, we are to determine the theoretical yield of iron metal

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Fe₂O₃ + 3CO → 2Fe + 3CO₂

This means

1 mole of iron(III) oxide reacts with 3 moles of carbon monoxide to produce 2 moles of iron metal and 3 moles of carbon dioxide gas.

Now, to determine the theoretical yield of iron metal

First, we will determine the number of moles of each reactant present

• For iron(III) oxide

Mass = 75.0 grams

Molar mass = 159.69 g/mol

Using the formula,

$Number \ of \ moles = \frac{Mass}{Molar\ mass}$

Then,

Number of mole of Fe₂O₃ present = $\frac{75.0}{159.69}$

Number of mole of Fe₂O₃ present = 0.46966 mole

• For carbon monoxide

Mass = 50.0 g

Molar mass = 44.01 g

Then,

Number of moles of CO₂ present = $\frac{50.0}{44.01}$

Number of moles of CO₂ present = 1.136105 moles

Now,

If 1 mole of iron(III) oxide reacts with 3 moles of carbon monoxide to produce 2 moles of iron metal

Then,

0.378702 mole of iron(III) oxide will react with 1.136105 moles of carbon monoxide to produce 0.757403 moles of iron metal

Therefore, the number of moles of iron metal that would be produced is 0.757403 moles

Now, for the mass of the iron metal

Using the formula

Mass = Number of moles × Molar mass

Atomic mass of iron metal = 55.845 g/mol

Then,

Mass of iron metal that would be produced = 0.757403 × 55.845

Mass of iron metal that would be produced = 42.29717 g

Mass of iron metal that would be produced ≅ 42.3 g

Hence, the theoretical yield of iron metal is 42.3 g

The equation for the reaction is:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Since, all of the carbon monoxide was consumed during the reaction, then carbon monoxide is the limiting reactant

Thus, iron (III) oxide is the excess reactant

Learn more on Stoichiometry here: brainly.com/question/6061451