Question

How many grams of Fe2O3 are produced from 6.3g of O3?

Answer 1

To determine the amount of Fe2O3 (iron(III) oxide) produced from a given amount of O3 (ozone), you need to write a balanced chemical equation for the reaction and then use stoichiometry to find the corresponding amount of Fe2O3. 454 grams of Fe2O3 are produced from 6.3g of O3.

Assuming a balanced chemical equation for the reaction is:

`\[4 \text{Fe} + 3 \text{O}_3 \rightarrow 2 \text{Fe}_2\text{O}_3\]`

This equation implies that four moles of iron react with three moles of ozone to produce two moles of iron(III) oxide.

Step 1. Determine the molar mass of O3:

- Oxygen (O) has a molar mass of approximately 16 g/mol.

- Ozone (O3) has three oxygen atoms, so the molar mass of O3 is
`\(3 * 16\)` g/mol.

Step 2. Calculate the moles of O3:

`\[\text{moles of O3} = \frac{\text{mass of O3}}{\text{molar mass of O3}}\]`

Step 3. Use the mole ratio from the balanced equation to find moles of Fe2O3:

`\[\text{moles of Fe}_2\text{O}_3 = \frac{\text{moles of O3}}{3} * 2\]`

Step 4. Calculate the mass of Fe2O3:

`\[\text{mass of Fe}_2\text{O}_3 = \text{moles of Fe}_2\text{O}_3 * \text{molar mass of Fe}_2\text{O}_3\]`

Let's calculate:

1. Molar mass of O3:

`\[\text{Molar mass of O3} = 3 * 16 \, \text{g/mol} = 48 \, \text{g/mol}\]`

2. Moles of O3:

`\[\text{moles of O3} = \frac{6.3 \, \text{g}}{48 \, \text{g/mol}} \approx 0.13125 \, \text{mol}\]`

3. Moles of Fe2O3:

`\[\text{moles of Fe}_2\text{O}_3 = \frac{0.13125 \, \text{mol}}{3} * 2 \approx 0.0875 \, \text{mol}\]`

4 Mass of Fe2O3:

`\[\text{mass of Fe}_2\text{O}_3 = 0.0875 \, \text{mol} * \text{molar mass of Fe}_2\text{O}_3\]`

The molar mass of Fe2O3 can be calculated by summing the molar masses of two iron atoms and three oxygen atoms:
`\[\text{Molar mass of Fe}_2\text{O}_3 = 2 * \text{molar mass of Fe} + 3 * \text{molar mass of O} = 2 * 55.85 + 3 * 16\]`

=159.7 g/mol