How much heat is released if 2.00 moles
CF2Cl2 reacts with 70.0 g F2 with a 75.0
percent yield?
CF2Cl2 + F2 → CF4 + Cl2
∆H for this reaction is -401 kJ/mol rxn.
Well, you should never mix chemicals without adult supervision, unless you're making a volcano for your school project! Now, let me calculate some heat and spice up the science.
First, we need to find the limiting reactant, the one that gets used up completely. Since we have 2.00 moles of CF2Cl2 and 70.0 g of F2, we need to convert the grams of F2 to moles. The molar mass of F2 is about 38.0 g/mol, so 70.0 g of F2 would be about 1.84 moles.
Since CF2Cl2 has a 2:1 ratio with F2, we can see that F2 is in excess. It's like going to a buffet and taking seconds when you don't need to.
Now, let's calculate the moles of CF2Cl2 that react. Since F2 is in excess, we can take the moles of F2 (1.84) and divide it by the coefficient of F2 (1), which gives us 1.84 moles of CF2Cl2 reacting.
Next, we multiply that by the molar enthalpy change (∆H) of the reaction, which is -401 kJ/mol rxn. So, -401 kJ/mol rxn times 1.84 moles of CF2Cl2 gives us -737.84 kJ of heat released.
But wait! We have to consider the 75.0 percent yield. It's like baking cookies and only getting 75% of the cookies you expected. So, we multiply -737.84 kJ by 75.0% (or 0.75) to get -553.38 kJ of heat released.
So, the amount of heat released if everything goes according to plan is approximately -553.38 kJ. Remember, though, whenever dealing with chemicals, safety first!
To calculate the heat released in this reaction, we need to consider the stoichiometry of the reaction and the yield.
First, let's determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines how much product is formed. We can compare the moles of CF2Cl2 and F2 to determine this.
Given:
CF2Cl2: 2.00 moles
F2: 70.0 g
To find the moles of F2, we can use its molar mass:
Molar mass of F2 = 19.00 g/mol
Moles of F2 = mass of F2 / molar mass of F2
Moles of F2 = 70.0 g / 19.00 g/mol
Moles of F2 = 3.68 moles
Now we can compare the moles of CF2Cl2 and F2:
CF2Cl2: 2.00 moles
F2: 3.68 moles
From the reaction equation, we can see that 1 mole of CF2Cl2 reacts with 1 mole of F2 to produce 1 mole of CF4. This means that the mole ratio of CF2Cl2 to F2 is 1:1. Since we have more moles of F2 (3.68 mol) compared to CF2Cl2 (2.00 mol), CF2Cl2 is the limiting reactant.
The balanced equation shows that for 1 mole of CF2Cl2, the reaction produces 401 kJ of heat (ΔH = -401 kJ/mol). Therefore, for 2.00 moles of CF2Cl2, the heat released would be 2.00 moles x -401 kJ/mol = -802 kJ.
However, we're given that the reaction has a 75.0 percent yield. So we need to calculate the actual heat released taking into account the yield.
Actual heat released = Theoretical heat released x Yield
Actual heat released = -802 kJ x 0.75
Actual heat released = -601.5 kJ
Therefore, if 2.00 moles of CF2Cl2 react with 70.0 g of F2 with a 75.0 percent yield, the amount of heat released is -601.5 kJ.
To calculate the amount of heat released in this reaction, we need to follow these steps:
Step 1: Write and balance the equation for the reaction.
The balanced equation for the reaction is:
CF2Cl2 + F2 → CF4 + Cl2
Step 2: Determine the limiting reactant.
To determine the limiting reactant, we need to compare the mole ratio between the reactants (CF2Cl2 and F2) and the given quantities.
From the balanced equation, we can see that 1 mole of CF2Cl2 reacts with 1 mole of F2 to produce 1 mole of CF4 and 1 mole of Cl2.
Given:
- 2.00 moles of CF2Cl2
- 70.0 g of F2
To find the moles of F2, we need to convert grams to moles. The molar mass of F2 is 38.0 g/mol.
Moles of F2 = Mass of F2 (g) / Molar mass of F2 (g/mol)
Moles of F2 = 70.0 g / 38.0 g/mol ≈ 1.842 moles
Comparing the mole ratio, we have an excess amount of F2 since we have 1.842 moles of F2 and only 2.00 moles of CF2Cl2. Therefore, CF2Cl2 is the limiting reactant.
Step 3: Calculate the moles of CF2Cl2 that reacted.
Since CF2Cl2 is the limiting reactant, we know that all 2.00 moles of CF2Cl2 reacted.
Step 4: Calculate the heat released.
The ∆H for the reaction is -401 kJ/mol rxn, which means that for every mole of the reaction that occurs, 401 kJ of heat is released.
So, the heat released in this reaction is calculated as follows:
Heat released = ∆H x moles of reaction
Heat released = -401 kJ/mol x 2.00 moles
Heat released = -802 kJ
The negative sign indicates that heat is released in the reaction.
However, you also mentioned a 75.0 percent yield. If the yield is given as a percentage, it means that only 75.0% of the theoretical amount of product is obtained. In that case, we need to multiply the heat released by the percentage yield.
Actual heat released = Heat released x Yield
Actual heat released = -802 kJ x (75.0/100)
Actual heat released = -601.5 kJ
Therefore, if 2.00 moles of CF2Cl2 reacts with 70.0 g of F2, with a 75.0% yield, the amount of heat released would be approximately -601.5 kJ.
First we must determine the limiting reagent(LR).
mols F2 = approx 70/38 = about 1.84.
2.0 mols CF2Cl2 will produce 2 mols CF4.
1.84 mols F2 will produce 1.84 mols CF4,
Therefore, the LR is F2 and CF2Cl2 is in excess.
So the rxn releases 401 kJ for 38 g F2. You have 70 g; therefore,
401 kJ x 70/38 = ? kJ released.