If 2.68 grams of Copper was recovered and you started with 8.94 grams of Copper(II)chloride,then what is the percent yield of the reaction?
8.94g CuCl2 x (atomic mass Cu/molar mass CuCl2) = ? g Cu = theoretical yield.
%yield = (actual yield/theoretical yield)*100 = ?
I don't understand it.Please help me!
You don't understand what part?
You start with 8.94 g CuCl2. You can obtain, if everything goes right and you have 100% on results, about 4.22 g. You recovered only 2.68g.
% yield or % recovery = (2.68/4.22)*100 = ?
To calculate the percent yield of a reaction, you need to compare the actual yield (the amount of product obtained in the experiment) to the theoretical yield (the maximum amount of product that could be obtained).
To find the theoretical yield, you need to use stoichiometry. The balanced chemical equation for the reaction will provide you with the mole ratios between the reactants and the product. In this case, the balanced equation is:
2 CuCl2 + Mg → 2 Cu + MgCl2
According to the balanced equation, 2 moles of CuCl2 react to produce 2 moles of Cu. Therefore, the molar ratio between CuCl2 and Cu is 1:1.
First, convert the mass of Copper(II) chloride (CuCl2) to moles using its molar mass. The molar mass of CuCl2 is 134.45 g/mol (63.55 g/mol for Cu + 2 * 35.45 g/mol for Cl). Thus:
8.94 g CuCl2 * (1 mol CuCl2 / 134.45 g CuCl2) = 0.0664 mol CuCl2
Since the molar ratio between CuCl2 and Cu is 1:1, the theoretical yield of Cu is also 0.0664 mol.
Now, convert the theoretical yield from moles to grams using the molar mass of Cu, which is 63.55 g/mol:
0.0664 mol Cu * (63.55 g Cu / 1 mol Cu) = 4.22 g Cu (theoretical yield)
The actual yield of Cu obtained in the experiment is given as 2.68 g.
Finally, calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100:
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (2.68 g / 4.22 g) * 100 = 63.5%
Therefore, the percent yield of the reaction is 63.5%.