Question

How many grams of aluminum must be placed in an excess amount of copper (II) chloride solution in order to produce 4g of copper metal?

Answer 1

Answer:

let i know your answer in anhour

Answer 2

Answer:

1.13g

Explanation:

2Al + 3CuCl2 = 2AlCl3 + 3Cu

now 4g copper means 4/63.5 mole in 3Cu

2 part Al = 3 part Cu

2 mole Al = 3 mole Cu

now 2/3 mole Al = 1 mole Cu

hence 2/3× 4/63.5 mole Al = 4/63.5 mole Cu

8/190.5 = x/27 ( X is given mass of Al )

now x = 1.13g

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