For the following reaction 563 grams of carbon tetrachloride

For the following reaction, 5.63 grams of carbon tetrachloride are mixed with excess methane (CH₄) . The reaction yields 5.09 grams of dichloromethane (CH₂Cl₂) .

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

Solution

The balanced equation is

CH4 + CCl4 ----> 2 CH₂Cl₂

Number of moles of CCl4 = 5.63 g / 153.82 g/mol = 0.0366 mole

from the balanced equation we can say that

1 mole of CCl4 produces 2 mole of CH2Cl2 so

0.0366 mole of CCl4 will produce

= 0.0366 mole of CCl4 *(2 mole of CH2Cl2 / 1 mole of CCl4 )

= 0.0732 mole of CH2Cl2

mass of 1 mole of CH2Cl2 = 84.93 g

so the mass of 0.0732 mole of CH2Cl2 = 6.22 g

Therefore, theoretical yield of CH2Cl2 = 6.22 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (5.09 / 6.22)*100 = 81.8 %

Therefore, precent yield = 81.8