1. How many moles of copper could you obtain from the decomposition of 1.00mol of CuCO3*Cu(OH)2?...
1. How many moles of copper could you obtain from the decomposition of 1.00mol of CuCO3*Cu(OH)2?
2. Your calculated yiled for a particular precipitation reaction is 4.569g. After performing the experiment, you discover your actual yield is 4.354g. What is your percent recovery (percent yield) for the reaction?
Homework Answers
1) for 1mole of CuCO3*Cu(OH)2 ......................there are 2 moles of Cu
2) % of yield = (actual yield/theoritical yield)*100
= (4.354/4.569)*100 = 95.29% or 95.3 %
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