Answer:
The limiting reactant is CH₄
26.0g of CH₂Cl₂ is the maximum amount that can be formed
4.15g CCl₄ will remain
Explanation:
The reaction of methane, CH₄, with carbon tetrachloride, CCl₄ is:
CH₄ + CCl₄ → 2CH₂Cl₂
To find the maximum mass of dichloromethane that can be determined we need to find moles of methane and carbon tetrachloride:
Moles CH₄:
2.45g * (1mol / 16.04g) = 0.153 moles
Moles CCl₄:
27.7g * (1mol / 153.82g) = 0.180 moles
That means just 0.153 moles of CCl₄ will react until CH₄ is over.
The limiting reactant is CH₄
Assuming the whole 0.153 moles will react, the moles of CH₂Cl₂ will be:
0.153 moles CH₄ * (2 moles CH₂Cl₂ / 1 mole CH₄) = 0.306 moles of CH₂Cl₂
The mass is (Molar mass dichloromethane: 84.93g/mol):
0.306 moles of CH₂Cl₂ * (84.93g / mol) = 26.0g of CH₂Cl₂
The moles of CCl₄ that remain are:
0.180 moles - 0.153 moles = 0.027 moles
In grams:
0.027 moles * (153.82g / mol) = 4.15g CCl₄
