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

How many grams of CCL4 were formed? 116.9 g

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

Step-by-step 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`