You are asked to make 43 grams of iron (Fe) from iron III oxide (Fe2O3) and carbon monoxide (CO) as shown in the chemical equation below.

Fe2O3 + 3CO → 2Fe + 3CO2

How many grams of iron III oxide must you use?

Question 10 options:

61.49 g

122.99 g

0.77 g

6863 g

we need an answer bud

To determine the number of grams of iron III oxide (Fe2O3) required, we need to use stoichiometry and the molar masses of the compounds involved.

1. Find the molar mass of Fe2O3:
Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O)
= (2 x 55.845 g/mol) + (3 x 16.00 g/mol)
= 111.69 g/mol + 48 g/mol
= 159.69 g/mol

2. Use the stoichiometric ratio from the balanced equation:
According to the balanced equation, 1 mole of Fe2O3 reacts with 3 moles of CO to produce 2 moles of Fe.

3. Convert grams of Fe to moles of Fe:
The molar mass of Fe is 55.845 g/mol, so 43 g of Fe is equal to:
43 g / 55.845 g/mol = 0.77067 mol Fe

4. Apply stoichiometry to find the moles of Fe2O3:
From the stoichiometric ratio in the balanced equation, we know that 2 moles of Fe corresponds to 1 mole of Fe2O3.
Therefore, the moles of Fe2O3 required will be half (0.77067 / 2) of the moles of Fe.

5. Calculate the grams of Fe2O3:
The grams of Fe2O3 is equal to the moles of Fe2O3 multiplied by its molar mass:
Grams of Fe2O3 = (0.77067 mol Fe / 2) x (159.69 g/mol Fe2O3)
= 0.38534 x 159.69 g
≈ 61.49 g

Therefore, you need approximately 61.49 grams of iron III oxide (Fe2O3) to make 43 grams of iron (Fe). So the correct answer is 61.49 g.

To determine the grams of iron III oxide (Fe2O3) needed, we need to use stoichiometry, which is the calculation of quantities in chemical reactions.

Let's start by finding the molar mass of Fe2O3.

Fe2O3 has a molar mass of:
2(Fe) + 3(O) = 2 x 55.845 g/mol + 3 x 16.00 g/mol = 159.69 g/mol

Now we need to determine the molar mass of CO.

CO has a molar mass of:
C + O = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol

The stoichiometric ratio in the balanced chemical equation is 1:3. This means that for every 1 mole of Fe2O3, we need 3 moles of CO to produce 2 moles of Fe.

Now we can use the given mass of iron (43 grams) and the molar masses to find the moles of iron.

Formula: Moles = Mass/Molar mass

Moles of Fe = 43 g / 55.845 g/mol (molar mass of Fe) = 0.770 moles

Since the stoichiometric ratio is 1:2 between Fe2O3 and Fe, the number of moles of Fe2O3 needed will be half of the number of moles of Fe.

Moles of Fe2O3 = 0.770 moles / 2 = 0.385 moles

Now we can find the grams of Fe2O3 by multiplying the moles by the molar mass.

Grams of Fe2O3 = 0.385 moles x 159.69 g/mol (molar mass of Fe2O3) = 61.49 grams

Therefore, to make 43 grams of iron, you need to use 61.49 grams of iron III oxide (Fe2O3).

The correct option is 61.49 grams.

Fe2O3 + 3CO → 2Fe + 3CO2

moles Fe you want = grams/atomic mass = ?
Using the coefficients in the balanced equation, convert mols Fe you want to mols Fe2O3 you need. That is mols Fe from above x (1 mole Fe2O3/2 mols Fe) = ?
Now convert mols Fe2O3 to grams Fe2O3. g = mols x molar mass = ?
Post our work if you get stuck.