Answer:
28.69 g of CH₂Cl₂
Step-by-step 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.