Answer:
[tex]Y=84.5\%[/tex]
Explanation:
Hello,
In this case, given the reaction, we can compute the moles of chlorine that are consumed and the yielded grams of 1,1,2,2-tetrachloroethane by using their molar masses, 70.9 g/mol and 167.8 g/mol respectively by using their 2:1 molar ratio:
[tex]m_{C_2H_2Cl_4 }=15.0gCl_2*\frac{1molCl_2}{70.9gCl_2}*\frac{1molC_2H_2Cl_4 }{2molCl_2} *\frac{167.8gC_2H_2Cl_4 }{1molC_2H_2Cl_4 } \\\\m_{C_2H_2Cl_4 }=17.75gC_2H_2Cl_4[/tex]
Then we compute the percent yield:
[tex]Y=\frac{15.0g}{17.75g}*100\%\\ \\Y=84.5\%[/tex]
Best regards.
Answer:
94.6%
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
2Cl2(g) + C2H2(g) → C2H2Cl4(l)
Step 2:
Determination of the mass of Cl2 that reacted and the mass of C2H2Cl4 produced from the balanced equation.
This is illustrated below:
Molar Mass of Cl2 = 2 x 35.5 = 71g/mol
Mass of Cl2 from the balanced equation = 2 x 71 = 142g
Molar Mass of C2H2Cl4 = (12x2) + (2x1) + (35.5x4) = 168g.
Mass of C2H2Cl4 from the balanced equation = 1 x 168 = 168g
Therefore, 146g of Cl2 reacted and 168g of C2H2Cl4 were produced from the balanced equation.
Step 3:
Determination of the theoretical yield of C2H2Cl4.
This is illustrated below:
From the balanced equation above,
146g of Cl2 reacted to produce 168g of C2H2Cl4.
Therefore, 15g of Cl2 will react to produce = (15 x 168)/142 = 17.75g of C2H2Cl4.
Therefore, the theoretical yield of C2H2Cl4 is 17.75g
Step 4:
Determination of the percentage yield of C2H2Cl4.
The percentage yield of C2H2Cl4 can be obtained as follow:
Actual yield of C2H2Cl4 = 16.8g
Theoretical yield of C2H2Cl4 = 17.75g
Percentage yield of C2H2Cl4 =..?
Percentage yield = Actual yield /Theoretical yield x 100
Percentage yield = 16.8/17.75 x 100
Percentage yield = 94.6%
Therefore, the percentage yield of C2H2Cl4 is 94.6%
