Final answer:
To calculate the amount of excess reactant left over, we first determine the limiting reactant and the moles of the products formed. From there, we can calculate the moles of excess reactant consumed and the mass of the excess reactant remaining. In this case, 2.42g of CS₂ is left over when 6.00g of CS₂ gas reacts with 10.0g of Cl₂ gas.
Step-by-step explanation:
To determine the amount of excess reactant left over, we first need to calculate the amount of the limiting reactant consumed in the reaction. The reaction of interest is CS₂(g) + 3 Cl₂(g) → CCl₄(l) + S₂Cl₂(l). We will use the amount of the limiting reactant, Cl₂, to calculate the amount of CCl₄ formed.
First, let's calculate the moles of Cl₂ using its molar mass: 10.0 g Cl₂ * (1 mol Cl₂ / 70.9 g Cl₂) = 0.141 mol Cl₂.
Next, we use the stoichiometry of the reaction to determine the moles of CCl₄ formed. From the balanced equation, we can see that 1 mole of Cl₂ reacts to form 1 mole of CCl₄. Therefore, the moles of CCl₄ formed is also 0.141 mol.
Finally, we calculate the mass of the excess reactant, CS₂, using its molar mass: 6.00 g CS₂ * (1 mol CS₂ / 76.1 g CS₂) = 0.079 mol CS₂.
Since the reaction uses 0.141 mol of Cl₂ and 0.079 mol of CS₂, we can see that CS₂ is the limiting reactant because it is consumed completely, while Cl₂ is in excess.
Now, let's calculate the mass of the excess reactant remaining.
Since 0.141 mol of Cl₂ reacts to form 0.141 mol of CCl₄, we can use the stoichiometry of the reaction to determine the moles of CS₂ needed to react with 0.141 mol of Cl₂:
0.141 mol Cl₂ * (1 mol CS₂ / 3 mol Cl₂) = 0.047 mol CS₂.
The amount of excess CS₂ remaining is the difference between the moles of CS₂ initially present and the moles of CS₂ needed to react with the limiting reactant:
0.079 mol CS₂ - 0.047 mol CS₂ = 0.032 mol CS₂
Finally, we calculate the mass of the excess reactant remaining:
0.032 mol CS₂ * (76.1 g CS₂ / 1 mol CS₂) = 2.42 g CS₂.