Question

For the following reaction, 22.0 grams of iron are allowed to react with with 10.6 grams of oxygen gas.

iron (s) + oxygen (g) iron(II) oxide (s)

What is the maximum amount of iron(II) oxide that can be formed?

Answer 1

For the following reaction, 22.0 grams of iron are allowed to react with 10.6 grams of oxygen gas.

$$\text{iron (s)} + \text{oxygen (g)} \cdot \text{iron(II)} \text{ oxide (s)}$$

What is the maximum amount of iron(II) oxide that can be formed?

Solution:

$$2 \mathrm{Fe}_{255.8} + \mathrm{O}_{2} = 2 \mathrm{FeO}$$

For the equation of reaction and basic data determine, what component is taken in plenty. For that find number moles of each component:

$$\begin{array}{l} n(\mathrm{Fe}) = \dfrac{22.0}{2 \cdot 55.8} = 0.197 \text{ mole}; \\ n(\mathrm{O}_{2}) = \dfrac{10.6}{32.0} = 0.331 \text{ mole}. \end{array}$$

So, oxygen is taken in plenty. Therefore, the amount of iron(II) oxide determined from the mass of iron:

$$2 \mathrm{Fe}_{255.8} + \mathrm{O}_{2} = 2 \mathrm{FeO}$$

From the reaction, the mass of iron is:

$$X = \dfrac{22.0 \cdot 2 \cdot 71.8}{2 \cdot 55.8} = 28.3 \text{ g}.$$

Answer:

The maximum amount of iron(II) oxide (FeO) is 28.3 g.