Question

It is desired to produce 2.25 grams of dichloromethane (CH2Cl2) by the following reaction. If the percent yield of dichloromethane (CH2Cl2) is 65.5 %, how many grams of carbon tetrachloride would need to be reacted?
______grams carbon tetrachloride

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

Answer 1

Answer:- 3.12 g carbon tetrachloride are needed.

Solution:- The balanced equation is:

$CH_4+CCl_4\rightarrow 2CH_2Cl_2$

From given actual yield and percent yield we will calculate the theoretical yield that would be further used to calculate the grams of carbon tetrachloride.

percent yield formula is:

percent yield = $(\frac{actual}{theoretical})100$

$65.5=(\frac{2.25}{theoretical})100$

$theoretical=(\frac{2.25(100)}{65.5})$

theoretical = 3.44 g

From balanced equation, there is 2:1 mol ratio between dichloethane and carbon tetrachloride.

Molar mass of dichloroethane is 84.93 gram per mol and molar mass of carbon tetrachloride is 153.82 gram per mol.

$3.44gCH_2Cl_2(\frac{1molCH_2Cl_2}{84.93gCH_2Cl_2})(\frac{1molCCl_4}{2molCH_2Cl_2})(\frac{153.82gCCl_4}{1molCCl_4})$

= $3.12gCCl_4$

So, 3.12 grams of carbon tetrachloride are needed to be reacted.