Answer:
About 16.1 grams of oxygen gas.
Explanation:
The reaction between magnesium and oxygen can be described by the equation:
[tex]\displaystyle 2\text{Mg} + \text{O$_2$} \longrightarrow 2\text{MgO}[/tex]
24.4 grams of Mg reacted with Oâ‚‚ to produce 40.5 grams of MgO. We want to determine the mass of Oâ‚‚ in the chemical change.
Compute using stoichiometry. From the equation, we know that two moles of MgO is produced from every one mole of Oâ‚‚. Therefore, we can:
- Convert grams of MgO to moles of MgO.
- Moles of MgO to moles of Oâ‚‚
- And moles of Oâ‚‚ to grams of Oâ‚‚.
The molecular weights of MgO and Oâ‚‚ are 40.31 g/mol and 32.00 g/mol, respectively.
Dimensional analysis:
[tex]\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$_2$}[/tex]
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:
[tex]\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$_2$}[/tex]
