Question

For the following reaction, 2.69 grams of methane (CH4) are allowed to react with 30.5 grams of carbon tetrachloride. methane (CH4) (g) + carbon tetrachloride (g) dichloromethane (CH2Cl2) (g) What is the maximum amount of dichloromethane (CH2Cl2) that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit AnswerRetry Entire Group6 more group attempts remaining

Answer 1

Answer: a) The maximum amount of dichloromethane $(CH_2Cl_2)$ that can be formed is 29 grams.

b) The formula for the limiting reagent is $CH_4$

c) Amount of the excess reagent i.e $CCl_4$ remains after the reaction is complete is 0.02 moles.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $CH_4$

$\text{Number of moles}=\frac{2.69g}{16g/mol}=0.17moles$

b) moles of $CCl_4$

$\text{Number of moles}=\frac{30.5g}{153.9g/mol}=0.19moles$

The balanced chemical equation will be:

$CH_4+CCl_4\rightarrow 2CH_2Cl_2$

According to stoichiometry :

1 mole of $CH_4$ require 1 mole of $CCl_4$

Thus 0.17 mole of $CH_4$ require=$\frac{1}{1}\times 0.17=0.17moles$  of $CCl_4$

Thus $CH_4$ is the limiting reagent as it limits the formation of product and $CCl_4$ is the excess reagent as (0.19-0.17) =0.02 moles of $CCl_4$ are left unreacted.

As 1 moles of $CH_4$ give =  2 moles of $CH_2Cl_2$

Thus 0.17 moles of $CH_4$ give =$\frac{2}{1}\times 0.17=0.34moles$  of $CH_2Cl_2$

Mass of $CH_2Cl_2=moles\times {\text {Molar mass}}=0.34moles\times 84.93g/mol=29g$

Thus 29 g of $CH_2Cl_2$ will be produced from the given masses of both reactants.