What is the maximum mass of dichloromethane (CH2Cl2) that can be formed? grams

Question

For the following reaction, 2.46 grams of methane (CH₄) are allowed to react with 26.9 grams of carbon tetrachloride .

methane (CH₄)(g) + carbon tetrachloride(g)  dichloromethane (CH₂Cl₂)(g)

What is the maximum mass of dichloromethane (CH₂Cl₂) that can be formed?  IN GRAMS

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?  IN GRAMS

Answer 1

Balance reaction CH₄ + CCl₄ --> 2 CH₂Cl₂.

(2.46g CH₄) * (1 mol CH₄ / 16.04g CH₄) * (2 mol CH₂Cl₂ /1 mol CH₄) = 0.3067 mol CH₂Cl₂

(26.9g CCl₄) * (1 mol CCl₄ / 153.8g CCl₄) * (2 mol CH₂Cl₂ /1 mol CCl₄) = 0.3498 mol CH₂Cl₂

Formula CH₄ is the limiting reactant, producing (0.3067 mol CH₂Cl₂) * (84.93g CH₂Cl₂ /1 mol CH₂Cl₂) = 26.0g CH₂Cl₂.

Excess 0.0431 mol CCl₄ * (153.8g CCl₄ /1 mol CCl₄) = 6.62g CCl₄ remaining.