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?
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
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 gD) Wt of Al consumed = 26.98*1.48*2/3 = 26.62 gTotal amount of leftover of Al = 200-26.62 =173.38 gWt 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
= 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
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