Given the balanced chemical reaction expressed as:
[tex]2Au_2O_3\rightarrow4Au+3O_2[/tex]A Redox reaction is a reaction that involves the transfer of electrons and changes in the oxidation state between the elements.
The given chemical equation is therefore an oxidation-reduction reaction since it involves a change in the oxidation state of the elements.
Determine the moles of Au₂O₃
[tex]\begin{gathered} \text{Mole = }\frac{Mass}{Molar\text{ mass}} \\ \text{Mole of Au}_2O_3=\frac{10g}{441.93g\text{/mol}} \\ \text{Mole of Au}_2O_3=0.02263\text{moles} \end{gathered}[/tex]According to stochiometry, you can see that 2 moles of Gold(III)oxide produce 4 moles of Gold. Hence the moles of Gold produced will be:
[tex]\begin{gathered} \text{moles of Gold=}\frac{0.02263\times4}{2} \\ \text{moles of Gold=}0.02263\times2 \\ \text{moles of Gold=}0.0453\text{moles} \end{gathered}[/tex]Determine the mass of Gold.
[tex]\begin{gathered} \text{Mass of Gold=moles}\times molar\text{ mass} \\ \text{Mass of Gold=}0.0453\times196.97 \\ \text{Mass of Gold}=8.91\text{grams} \end{gathered}[/tex]Next is determining the mole of Oxygen
According to stochiometry, you can see that 2 moles of Gold(III)oxide produce 3 moles of Oxygen. Hence the moles of Oxygen produced will be:
[tex]\begin{gathered} \text{moles of Oxygen=}\frac{0.02263\times3}{2} \\ \text{moles of Oxygen}=0.033945\text{moles} \end{gathered}[/tex]Determine the mass of oxygen produced
[tex]\begin{gathered} \text{Mass of O}_2=moles\times\text{Molar mass} \\ \text{Mass of O}_2=0.033945\times16 \\ \text{Mass of O}_2=0.543\text{grams} \end{gathered}[/tex]Hence the mass of oxygen produced is 0.543 grams
