For the following reaction, 7.43 grams of methane (CH4) are allowed to react with 27.6 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

30.4 grams of CH2Cl2 will be formed

Explanation:

Step 1: data given

Mass of methane = 7.43 grams

Molar mass of methane (CH4) = 16.04 g/mol

Mass of CCl4 = 27.6 grams

Molar mass of CCl4 = 153.82 g/mol

Step 2: The balanced equation

CH4 + CCl4 → 2CH2Cl2

Step 3: Calculate moles

Moles = mass / molar mass

Moles methane = mass methane / molar mass methane

Moles methane = 7.43 grams / 16.04 g/mol

Moles methane = 0.463 moles

Moles CCl4 = 27.6 grams / 153.82 g/mol

Moles CCl4 = 0.179 moles

Step 4: Calculate limiting reactant

For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

CCl4 is the limiting reactant. It will completely be consumed. (0.179 moles). Ch4 is in excess. There will react 0.179 moles. There will remain 0.463 - 0.179 = 0.284 moles

Step 5: Calculate moles CH2Cl2

For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

For 0.179 moles CCl4 we'll have 2*0.179 = 0.358 moles CH2Cl2

Step 6: Calculate mass CH2Cl2

Mass CH2Cl2 = 0.358 moles * 84.93 g/mol

Mass CH2Cl2 = 30.4 grams

30.4 grams of CH2Cl2 will be formed