Consider the following balanced equation 2 Als 3 Cu Cl22 H2

Consider the following balanced equation 2 Al(s) + 3 Cu Cl2-2 H2O (aq) 3 Cu (s) + 2 AlCl3 (aq) + 6 H2O (D If we began the experiment with 0.65 g of CuCl2.2 H20, according to the stoichiometry of the reaction, how much Al, should be used to complete the reaction without either reactant being in excess? Answer the following using the correct number of significant figures.

Solution

Ans 1 :

Number of moles of CuCl2.2H2O = 0.65 / 170.48

= 0.0038 moles

3 moles of CuCl2.2H2O requires 2 mole of Al

So 0.0038 moles of CuCl2.2H2O will require : ( 0.0038 x 2) / 3

= 0.0025 mol of Al

So the mass of Al used in the reaction = 0.0025 x molar mass of Al

= 0.0025 x 26.98

= 0.067 grams