Question

4. Magnesium and oxygen undergo a chemical reaction to

produce magnesium oxide.
2Mg + O2 + 2MgO
If 24.4 grams of magnesium react with oxygen gas to produce
40.5 grams of magnesium oxide, what is the mass in grams of
the oxygen gas in the chemical change?

Answer 1

About 16.1 grams of oxygen gas.

Step-by-step explanation:

The reaction between magnesium and oxygen can be described by the equation:

`\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}`

24.4 grams of Mg reacted with O₂ to produce 40.5 grams of MgO. We want to determine the mass of O₂ in the chemical change.

Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of O₂. Therefore, we can:

1. Convert grams of MgO to moles of MgO.
2. Moles of MgO to moles of O₂
3. And moles of O₂ to grams of O₂.

The molecular weights of MgO and O₂ are 40.31 g/mol and 32.00 g/mol, respectively.

Dimensional analysis:

`\displaystyle 40.5\text{ g MgO} \cdot \frac{1\text{ mol MgO}}{40.31\text{ g MgO}} \cdot \frac{1\text{ mol O$_2$}}{2\text{ mol MgO}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}`

In conclusion, about 16.1 grams of oxygen gas was reacted.

You will obtain the same result if you compute with the 24.4 grams of Mg instead:

`\displaystyle 24.4\text{ g Mg}\cdot \frac{1\text{ mol Mg}}{24.31\text{ g Mg}} \cdot \frac{1\text{ mol O$_2$}}{1\text{ mol Mg}} \cdot \frac{32.00\text{ g O$_2$}}{1\text{ mol O$_2$}} = 16.1\text{ g O$_2$}`