A chemist is producing copper(II) chloride using the following reaction:
Cu(s) + Cl2(g) --> CuCl2(s)
In one reaction, the chemist uses 15.0 g of copper metal and 10.0 g of chlorine gas as reactants. When the reaction reaches completion, 16.0 g of copper(II) chloride has been produced. What was the percent yield of the reaction?
must be each reactant ?
or total reactants ?
Ill do each
Step one:
mm of products
cu=63.546
2cl=70.90
------------
134. 45g/mol
step two
mol of products
16.0 g/134.45=1.19 x 10^-1 mol x
step two
find the lr
cu=15.0 /63.546 =.236mol produced
cl=10.0/70.90=.141
lr is cl
three
turn it into grams
.141mol x 134. 45g/mol=18.96 g
four % yield
16.0g/18.96 g x 100=84.4% around.
To calculate the percent yield of a reaction, you need to compare the actual yield (the amount of copper(II) chloride produced) to the theoretical yield (the amount of copper(II) chloride that would be produced if the reaction went to completion).
First, let's calculate the theoretical yield of copper(II) chloride. We'll determine the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
1. Start with the given masses of reactants:
- Mass of copper (Cu): 15.0 g
- Mass of chlorine gas (Cl2): 10.0 g
2. Convert the masses of Cu and Cl2 to moles using their respective molar masses:
- Molar mass of Cu: 63.55 g/mol
- Molar mass of Cl2: 70.91 g/mol
Moles of Cu = mass of Cu / molar mass of Cu = 15.0 g / 63.55 g/mol = 0.236 mol
Moles of Cl2 = mass of Cl2 / molar mass of Cl2 = 10.0 g / 70.91 g/mol = 0.141 mol
3. Use the coefficients from the balanced chemical equation to find the mole ratio between Cu and CuCl2:
Cu(s) + Cl2(g) -> CuCl2(s)
According to the balanced equation, the mole ratio is 1:1. So, for every 1 mole of Cu, 1 mole of CuCl2 is produced.
Therefore, the moles of CuCl2 produced would be equal to the moles of Cu, which is 0.236 mol.
4. Calculate the molar mass of CuCl2:
- Molar mass of CuCl2: 134.45 g/mol
5. Calculate the theoretical yield of CuCl2 using the moles and molar mass:
Theoretical yield = moles of CuCl2 * molar mass of CuCl2
Theoretical yield = 0.236 mol * 134.45 g/mol = 31.74 g
Now that we have the theoretical yield, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) * 100%
Given: actual yield = 16.0 g (as stated in the question)
Theoretical yield = 31.74 g (calculated above)
Percent yield = (16.0 g / 31.74 g) * 100% ≈ 50.4%
Therefore, the percent yield of the reaction is approximately 50.4%.