For the following reaction, 2.70 grams of methane (CH4) are allowed to react with 30.0 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

28.69 g of CH₂Cl₂

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

CH₄ + CCl₄ —> 2CH₂Cl₂

Next, we shall determine the masses of CH₄ and CCl₄ that reacted and the mass of CH₂Cl₂ produced from the balanced equation. This is illustrated below:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4 = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of CCl₄ = 12 + (4×35.5)

= 12 + 142 = 154 g/mol

Mass of CCl₄ from the balanced equation = 1 × 154 = 154 g

Molar mass of CH₂Cl₂ = 12 + (2×1) + (2×35.5)

= 12 + 2 + 71 = 85 g/mol

Mass of CH₂Cl₂ from the balanced equation = 2 × 85 = 170 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 154 g of CCl₄ to produce 170 g of CH₂Cl₂.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted with 154 g of CCl₄.

Therefore, 2.7 g of CH₄ will react with = (2.7 × 154)/16 = 25.99 g of CCl₄

From the calculations made above, we can see that only 25.99 g of CCl₄ out of 30 g given is required to react completely with 2.7 g of CH₄.

Therefore, CH₄ is the limiting reactant and CCl₄ is the excess reactant.

Finally, we shall determine the maximum mass of CH₂Cl₂ produced from the reaction.

In this case, the limiting reactant will be used because it will produce the maximum mass of CH₂Cl₂ since all of it is consumed in the reaction.

The limiting reactant is CH₄ and the maximum mass of CH₂Cl₂ produce can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted to produce 170 g of CH₂Cl₂.

Therefore, 2.7 g of CH₄ will react to produce = (2.7 × 170)/16 = 28.69 g of CH₂Cl₂.

Thus, 28.69 g of CH₂Cl₂ were obtained from the reaction.

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