To produce 3.50 moles of silver, 1.75 moles of copper must react based on the balanced chemical equation provided. The stoichiometric ratio shows that 1 mole of Cu produces 2 moles of Ag. Therefore, the correct answer is 1.75 moles of Cu.
;
Analyze the balanced chemical equation: C u + 2 A g N O 3 → 2 A g + C u ( N O 3 ) 2 .
Determine the molar ratio between C u and A g : 1 mole of C u produces 2 moles of A g , so the ratio is 2 1 .
Set up a proportion to find the moles of C u needed to produce 3.50 moles of A g : 3.50 x = 2 1 .
Solve for x : x = 2 3.50 = 1.75 . The final answer is 1.75 mol.
Explanation
Problem Analysis We are given the balanced chemical equation for the reaction of copper ( C u ) and silver nitrate ( A g N O 3 ) : C u + 2 A g N O 3 → 2 A g + C u ( N O 3 ) 2 We are asked to find how many moles of copper ( C u ) must react to form 3.50 mol of A g .
Molar Ratio From the balanced equation, we can see that 1 mole of C u reacts to produce 2 moles of A g . This gives us the molar ratio: m o l es _ o f _ A g m o l es _ o f _ C u = 2 1 We can use this ratio to find the number of moles of C u needed to produce 3.50 moles of A g .
Solving for Moles of Copper Let x be the number of moles of C u needed. We can set up the following proportion: 3.50 x = 2 1 To solve for x , we multiply both sides of the equation by 3.50: x = 2 3.50 x = 1.75
Final Answer Therefore, 1.75 moles of copper must react to form 3.50 moles of silver.
Examples
In the field of jewelry making, understanding stoichiometry is crucial. For instance, if a jeweler wants to produce a certain amount of silver jewelry from a reaction involving copper and silver nitrate, they need to calculate the exact amount of copper required to react with silver nitrate to yield the desired amount of silver. This ensures that the reaction is efficient and that the jeweler obtains the desired quantity of silver without wasting materials. Stoichiometry helps in scaling up or down chemical reactions to meet specific production needs, making it an indispensable tool in manufacturing and chemical industries.