In a reaction between copper metal and aqueous iron(3) chloride, Fe^3+ is reduced to Fe^2+ (Aq), while Cu(s) is oxidized to Cu^2+(aq). Determine the mass of Cu(s) required to react with 42.8 mL of 1.61 mol/L FeCl3(aq)
To calculate the mass of Cu(s) required to react with 42.8 mL of 1.61 mol/L FeCl3(aq), we need to use the concept of stoichiometry.
1. Write a balanced chemical equation for the reaction:
2 FeCl3(aq) + 3 Cu(s) -> 3 CuCl2(aq) + 2 FeCl2(aq)
2. Determine the moles of FeCl3(aq) using the given volume and concentration:
Moles of FeCl3 = Volume (L) x Concentration (mol/L)
Moles of FeCl3 = 0.0428 L x 1.61 mol/L = 0.069088 mol
3. Use the stoichiometry of the balanced equation to find the moles of Cu(s):
From the balanced equation, we can see that the ratio of FeCl3 to Cu is 2:3.
Moles of Cu = (2/3) x Moles of FeCl3
Moles of Cu = (2/3) x 0.069088 mol = 0.04605867 mol
4. Calculate the molar mass of Cu:
The molar mass of Cu is 63.55 g/mol.
5. Calculate the mass of Cu(s) required:
Mass of Cu = Moles of Cu x Molar mass of Cu
Mass of Cu = 0.04605867 mol x 63.55 g/mol = 2.930 g
Therefore, the mass of Cu(s) required to react with 42.8 mL of 1.61 mol/L FeCl3(aq) is approximately 2.930 grams.