A reaction is run in which dissolved silver ions are reduced in the presence of solid copper. THis reaction can be used to produce a silver coating on a copper wire frame. .249g of copper wire was placed in a 100 mL of a 0.500 M solution of silver nitrate. What is the maximum quantity, in grams, of solid silver that could be recovered?
To find the maximum quantity of solid silver that could be recovered from the reaction, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Let's calculate the amount of silver that can be produced from the reaction:
1. Determine the moles of copper wire:
- Given mass of copper wire = 0.249 g
- Copper's molar mass = 63.55 g/mol
- Moles of copper = mass of copper wire / molar mass = 0.249 g / 63.55 g/mol
2. Calculate the moles of silver ions in the solution:
- Volume of silver nitrate solution = 100 mL = 0.100 L
- Silver nitrate concentration = 0.500 M
- Moles of silver ions = volume of solution * concentration = 0.100 L * 0.500 mol/L
3. Determine the stoichiometric ratio between copper and silver:
- From the balanced chemical equation: 2 Ag⁺ + Cu → Cu²⁺ + 2 Ag
- 2 moles of silver ions are required per 1 mole of copper (Cu).
4. Identify the limiting reagent:
- The limiting reagent is the one that has the smaller number of moles. Compare the moles of copper and silver ions calculated in steps 1 and 2.
5. Calculate the maximum moles of silver that could be obtained:
- Since copper is the limiting reagent, the moles of silver will depend on the moles of copper.
- Moles of silver = (2 moles of silver ions / 1 mole of copper) * moles of copper
6. Convert moles of silver to grams:
- Molar mass of silver = 107.87 g/mol
- Mass of silver = moles of silver * molar mass of silver
Now, let's calculate the maximum quantity of solid silver that could be recovered using these steps: