Based on the following equation: #2 Al + 3 CuCl_2 - 2AlCl_3 + 3 Cu#. lf 10.53 grams of #CuCl_2# were reacted, how many grams of #AlCl_3# will be produced?

Question

Based on the following equation: #2 Al + 3 CuCl_2 -> 2AlCl_3 + 3 Cu#. lf 10.53 grams of #CuCl_2# were reacted, how many grams of #AlCl_3# will be produced?

Owen Ambrose · Accepted Answer

The law of equivalent proportion may help solving the problem.

According to this law a reactant will produce same no of equivalent of product/s as its reacting mass has. here the equivalent mass of #CuCl_2# = Formula wt /total valency of metal(Cu) = #(63.55+2*35.55)/(1*2)# =67.33 Again the equivalent mass of #AlCl_3# = Formula wt /total valency of metal(Al) = #(27+3*35.55)/(1*3)# =44.55 Now no. of gram equivalent in 10.53g of #CuCl_2# reacted = #10.53/67.33#=0.156 hence the same no.of gram equivalent of #AlCl_3# will be produced So the mass of #AlCl_3# produced =0.156x equivalent mass of #AlCl_3# 0.156x44.55 = 6.95 g Please provide feedback.

Kai Bowman · Answer

undefined To find the grams of AlCl3 produced, you first need to determine the molar mass of CuCl2 and AlCl3. Then, use stoichiometry to find the amount of AlCl3 produced from the given amount of CuCl2.

1. Calculate the molar mass of CuCl2:
   Cu: 1 * 63.55 g/mol = 63.55 g/mol
   Cl: 2 * 35.45 g/mol = 70.90 g/mol
   Total: 63.55 + 70.90 = 134.45 g/mol

2. Calculate the molar mass of AlCl3:
   Al: 1 * 26.98 g/mol = 26.98 g/mol
   Cl: 3 * 35.45 g/mol = 106.35 g/mol
   Total: 26.98 + 106.35 = 133.33 g/mol

3. Calculate the moles of CuCl2:
   Moles = Mass / Molar mass
   Moles = 10.53 g / 134.45 g/mol ≈ 0.078 moles

4. Use the stoichiometry from the balanced equation to find the moles of AlCl3 produced:
   From the equation, 2 moles of AlCl3 are produced for every 3 moles of CuCl2.
   Moles of AlCl3 = (Moles of CuCl2 * 2) / 3
                  = (0.078 moles * 2) / 3 ≈ 0.052 moles

5. Calculate the mass of AlCl3 produced:
   Mass = Moles * Molar mass
   Mass = 0.052 moles * 133.33 g/mol ≈ 6.94 grams

Therefore, approximately 6.94 grams of AlCl3 will be produced.

Based on the following equation: #2 Al + 3 CuCl_2 -&gt; 2AlCl_3 + 3 Cu#. lf 10.53 grams of #CuCl_2# were reacted, how many grams of #AlCl_3# will be produced?

Based on the following equation: #2 Al + 3 CuCl_2 -> 2AlCl_3 + 3 Cu#. lf 10.53 grams of #CuCl_2# were reacted, how many grams of #AlCl_3# will be produced?