##### 2Al+3CuCl2=3Cu+2AlCl3

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A. Balance

B.If there are 200g of each reactant, which reactant is limiting?

C. What is the theoretical yield of each product?

D. How much excess reactant will be used? How much will be leftover?

E. If your percent yield was 74.4%, how many grams of solid metal product were collected?

Apr 22nd, 2015

A.) Balanced chemical reaction

2 Al + 3CuCl2  = 3Cu +  2AlCl3

B)Molecular wt of Al = 26.98 g/mol

Hence moles of Al = 200/26.98 = 7.41 moles

Molecular wt of CuCl2 = 134.45 g/mol

Thus moles of CuCl2 = 200/134.45 = 1.48 moles

Now for balanced chemical reaction

(Number of mole of CuCl2 = 3) > (number of moles of Al= 2)

But (number of moles of CuCl2=1.48) < (moles of Al = 7.41)

Hence CuCl2 is a limiting agent.

C) Number of moles of Cu produced = Number of moles of CuCl2 consumed = 1.48 moles

Molecular wt of Cu = 63.54 g/mol

Hence,

Theoretical yield of Copper = 1.48*63.54 = 94.03 g

Similarly theoretical yield of AlCl3 =molar wt of AlCl3 moles of CuCl3*2/3

= 1.48*133.34 *2/3 = 131.56 g

D)  Wt of Al consumed = 26.98*1.48*2/3 = 26.62 g

Total amount of leftover of Al = 200-26.62 =173.38 g

Wt of CuCl2 leftover = 0 (since limiting reactant)

E)Solid metal collected of Cu = moles of CuCl2*yield*molar mass of Cu

= 1.48*0.744*63.54

=69.96 g

Apr 22nd, 2015

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Apr 22nd, 2015
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Apr 22nd, 2015
May 25th, 2017
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