A scientist wants to determine how many moles of copper (II) oxide (CuO) would be needed to make 8.10 mol of gold (III) oxide [tex]$\left( Au _2 O _3\right)$[/tex].
[tex]$2 Au+3 CuO-Au_2 O_3+3 Cu$[/tex]
How many moles of copper (II) oxide would be needed?
[tex]$8.10 mol Au _2 O _3$[/tex] [$\square$]
[$\square$] mol CuO , [$\square$] mol CuO
[tex]$mol Au I _2 O _3$[/tex]
Answer (1)
Identify the stoichiometric ratio between C u O and A u 2 O 3 from the balanced equation.
Multiply the moles of A u 2 O 3 by the stoichiometric ratio to find the moles of C u O needed.
Perform the calculation: 8.10 m o l A u 2 O 3 × 1 m o l A u 2 O 3 3 m o l C u O = 24.3 m o l C u O .
State the final answer: 24.3
Explanation
Understanding the Problem We are given the balanced chemical equation: 2 A u + 3 C u O r i g h t a rro w A u 2 O 3 + 3 C u . We want to find out how many moles of copper (II) oxide ( C u O ) are needed to produce 8.10 moles of gold (III) oxide ( A u 2 O 3 ).
Finding the Stoichiometric Ratio From the balanced equation, we can see that 3 moles of C u O are required to produce 1 mole of A u 2 O 3 . This gives us the stoichiometric ratio: 1 m o l A u 2 O 3 3 m o l C u O
Calculating Moles of CuO To find the number of moles of C u O needed to produce 8.10 moles of A u 2 O 3 , we multiply the given moles of A u 2 O 3 by the stoichiometric ratio: m o l es o f C u O = 8.10 m o l A u 2 O 3 × 1 m o l A u 2 O 3 3 m o l C u O
Final Calculation m o l es o f C u O = 8.10 × 3 = 24.3 Therefore, 24.3 moles of C u O are needed to produce 8.10 moles of A u 2 O 3 .
Examples
In chemistry, stoichiometry is essential for calculating the amounts of reactants and products in chemical reactions. For instance, if you're synthesizing a new material and the reaction requires a specific ratio of ingredients, understanding stoichiometry ensures you add the correct amounts to achieve the desired product. In this case, knowing how many moles of copper (II) oxide are needed to produce a certain amount of gold (III) oxide is crucial for efficient and successful synthesis, preventing waste and optimizing the reaction yield. This principle applies to various fields, from pharmaceuticals to materials science, where precise chemical compositions are vital.
