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?

Answer 1

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

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Answer 2

Kai Bowman

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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?