Answer:
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
