Question

For the following reaction, 3.99 grams of carbon tetrachloride are mixed with excess methane (CH4) . The reaction yields 3.58 grams of dichloromethane (CH2Cl2) .

methane (CH4) ( g ) + carbon tetrachloride ( g ) dichloromethane (CH2Cl2) ( g )

What is the theoretical yield of dichloromethane (CH2Cl2) ? grams
What is the percent yield for this reaction ? %

Answer 1

1. The theoretical yield (in grams) of dichloromethane is 4.40 grams

2. The percent yield of dichloromethane is 81.2%

Calculation of theoretical yield and percent yield

1. The theoretical yield of dichloromethane can be calculated as shown below:

• 
  `CH_4\ +\ CCl_4\ \rightarrow\ 2CH_2Cl_2`
• Molar mass of
  `CCl_4` = 154 g/mol
• Mass of
  `CCl_4` from the balanced equation = 1 × 154 = 154 g
• Molar mass of
  `CH_2Cl_2` = 85 g/mol
• Mass of
  `CH_2Cl_2` from the balanced equation = 2 × 85 = 170 g
• Theoretical yield of
  `CH_2Cl_2` =?

From the balanced equation,

154 grams of
`CCl_4` reacted to form 170 grams of
`CH_2Cl_2`

Therefore,

3.99 grams of
`CCl_4` will react to form =
`(3.99\ *\ 170)/(154)` = 4.40 grams of
`CH_2Cl_2`

Thus, the theoretical yield of
`CH_2Cl_2` is 4.40 grams

2. The percent yield of dichloromethane can be calculated as follow:

• Actual yield of
  `CH_2Cl_2` = 3.58 grams
• Theoretical yield of
  `CH_2Cl_2` = 4.4- grams
• Percentage yield of
  `CH_2Cl_2` =?

`Percentage\ of\ CH_2Cl_2\ = (Actual\ yield)/(theoretical\ yield)\ *\ 100\\ \\= (3.58)/(4.4) *\ 100\\\\= 81.2\%`