Question

Copper(II) Chloride (CuCl2; 0.98 g) was dissolved in water and piece of aluminum wire (Al; 0.56 g) was placed in the solution. The blue color due to copper(II) chloride soon faded and a red precipitate of solid copper was observed. After the blue color had disappeared completely, the leftover aluminum wire was removed from the solution and weighed. The mass of the leftover aluminum wire was 0.43 g.
1. Calculate the number of moles of (a) copper(II) chloride and (b) aluminum that reacted.

Answer 1

To calculate the number of moles of copper(II) chloride, we use the equation:

moles = mass / molar mass

The molar mass of copper(II) chloride (CuCl2) can be found in the periodic table. The atomic masses of copper (Cu) and chlorine (Cl) are 63.55 g/mol and 35.45 g/mol, respectively.

(a) moles of copper(II) chloride = 0.98 g / (63.55 g/mol + 2(35.45 g/mol))
= 0.98 g / 134.45 g/mol
= 0.0073 mol

To calculate the number of moles of aluminum, we also use the equation:

moles = mass / molar mass

The molar mass of aluminum (Al) is 26.98 g/mol.

(b) moles of aluminum = 0.56 g / 26.98 g/mol
= 0.0208 mol