Question

Methane and chlorine react to form four products: CH3Cl, CH2Cl2, CHCl3, and CCl4. At a particular temperature and pressure, 38.4 g of CH4 was allowed to react with excess Cl2 and gave 9.2 g CH3Cl, 47.1 g CH2Cl2, and 109 g CHCl3. All the CH4 reacted. (Note: The hydrogen that is displaced from the carbon also combines with Cl2 to form HCl.)How many grams of CCL4 were formed?How many grams of Cl2 reacted with the CH4?

Answer 1

Answer:

How many grams of CCL4 were formed? 116.9 g

How many grams of Cl2 reacted with the CH4? 243.8 g

Explanation:

First we need to know the molar mass for every element or compound in the reaction:

$M_{CH_{4}}=16 g/mol\\M_{CH_{3}Cl}=50.49g/mol\\M_{CH_{2}Cl_{2}}=84.93g/mol\\M_{CHCl_{3}}=119.38g/mol\\M_{CCl_{4}}=153.82g/mol$

Now we proceed to calculate the amount of moles produced, per product:

$n_{CH_{4}}=2.4\\n_{CH_{3}Cl}=0.18\\n_{CH_{2}Cl_{2}}=0.55\\n_{CHCl_{3}}=0.91\\n_{CCl_{4}}=n_{CH_{4}}-(n_{CH_{3}Cl}+n_{CH_{2}Cl_{2}}+n_{CHCl_{3}})\\n_{CCl_{4}}=0.76mol\\m_{CCl_{4}}=n_{CCl_{4}}*M_{CCl_{4}}\\m_{CCl_{4}}=116.9g$

To calculate the mass of chlorine we just need to make a mass balance:

$m_{CH_{4}}+m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}\\m_{Cl_{2}}=m_{CH_{3}Cl}+m_{CH_{2}Cl_{2}}+m_{CHCl_{3}}+m_{CCl_{4}}-m_{CH_{4}}\\m_{Cl_{2}}=243.8g$