The overall reaction in a commercial heat pack can be represented as shown below.
4 Fe(s) + 3 O2(g) 2 Fe2O3(s) ΔH = -1652 kJ
How much heat is released when 4.00 mol iron is reacted with excess O2?
How much heat is released when 1.00 mol Fe2O3 is produced?
How much heat is released when 1.00 iron is reacted with excess O2?
How much heat is released when 10.0g Fe and 2.00g O2 are reacted?
I've been looking at this for a long time, and I'm really stuck on starting each question...
Well, don't worry! I'm here to help you out with a little humor.
Q1: How much heat is released when 4.00 mol iron is reacted with excess O2?
A1: Well, let's just say this reaction is hotter than a summer romance. With 4.00 mol of iron, the heat released can be calculated using the provided enthalpy change value of -1652 kJ.
Q2: How much heat is released when 1.00 mol Fe2O3 is produced?
A2: Ah, the magic of chemistry! When 1.00 mol of Fe2O3 is produced, the same enthalpy change of -1652 kJ can be used to calculate the heat released. It's like watching fireworks on the Fourth of July – a sight to behold!
Q3: How much heat is released when 1.00 iron is reacted with excess O2?
A3: Well, if 1.00 iron gets into a hot date with excess O2, things are bound to heat up! In this case, we can use the stoichiometry of the reaction and the provided enthalpy change to calculate the heat released. So, buckle up for some exothermic fun!
Q4: How much heat is released when 10.0g Fe and 2.00g O2 are reacted?
A4: Chemistry and cooking have something in common – heat! Speaking of which, when 10.0g of iron and 2.00g of O2 decide to have a little reaction party, we can use their respective molar masses to convert them into moles. From there, we can use stoichiometry and the enthalpy change to calculate the amount of heat released. This reaction is definitely cooking up a storm!
I hope these lighthearted explanations help you get started on those questions. Remember, chemistry can be a funny business!
To solve each question, you need to use the given balanced equation and the molar ratio between reactants and products. The molar ratio tells you how many moles of one substance reacts with another substance during a chemical reaction.
Let's go through each question step by step:
1. How much heat is released when 4.00 mol iron is reacted with excess O2?
To answer this question, we can use the stoichiometry from the balanced equation. The balanced equation tells us that 4 moles of Fe reacts to produce 2 moles of Fe2O3.
Since 2 moles of Fe2O3 release -1652 kJ of heat, we can set up the following proportion:
(2 mol Fe2O3 / -1652 kJ) = (4 mol Fe / x kJ)
Using this proportion, we can solve for x to find the amount of heat released when 4.00 mol of Fe is reacted.
2. How much heat is released when 1.00 mol Fe2O3 is produced?
Again, we can use the stoichiometry from the balanced equation. The balanced equation tells us that 4 moles of Fe reacts to produce 2 moles of Fe2O3.
Since 2 moles of Fe2O3 release -1652 kJ of heat, we can set up the following proportion:
(2 mol Fe2O3 / -1652 kJ) = (1 mol Fe2O3 / x kJ)
Using this proportion, we can solve for x to find the amount of heat released when 1.00 mol of Fe2O3 is produced.
3. How much heat is released when 1.00 mol iron is reacted with excess O2?
Similar to the previous questions, we can use the stoichiometry from the balanced equation. The balanced equation tells us that 4 moles of Fe reacts to produce 2 moles of Fe2O3.
Since 2 moles of Fe2O3 release -1652 kJ of heat, we can set up the following proportion:
(2 mol Fe2O3 / -1652 kJ) = (1 mol Fe / x kJ)
Using this proportion, we can solve for x to find the amount of heat released when 1.00 mol of Fe is reacted.
4. How much heat is released when 10.0g Fe and 2.00g O2 are reacted?
To answer this question, we need to calculate the number of moles for each reactant using their respective molar masses. The molar mass of Fe is given as 55.85 g/mol and the molar mass of O2 is 32.00 g/mol.
Using the molar masses, we can convert the given masses into moles:
10.0 g Fe * (1 mol Fe / 55.85 g Fe) = # of moles of Fe
2.00 g O2 * (1 mol O2 / 32.00 g O2) = # of moles of O2
Now that we know the moles of Fe and O2, we can use the stoichiometry from the balanced equation. The balanced equation tells us that 4 moles of Fe reacts with 3 moles of O2.
Using the stoichiometry, we can find the limiting reactant and determine the number of moles of Fe2O3 produced. Then, we can set up the proportion:
(2 mol Fe2O3 / -1652 kJ) = (# of moles of Fe2O3 / x kJ)
Using this proportion, we can solve for x to find the amount of heat released when 10.0g of Fe and 2.00g of O2 are reacted.
I hope this explanation helps you understand how to approach each question. Let me know if you need further assistance!
For your ***Last question, you have to determine the limiting reagent:
10.0g Fe*(1 mole/55.845g)= moles of Fe
2.00g O2*(1 mole/32.00g)= moles of O2
Eyeballing it, I think O2 is the limiting reagent.
Proceed as you would for the second problem.
This is just basic stoichiometry.
From the reaction, this is the basic information that you will need:
4 moles of Fe = -1652 kJ
3 moles of O2 = -1652 kJ
and
2 moles of Fe2O3= -1652 kJ
So, for your first question:
4 moles of Fe = -1652 kJ
For your second question:
1.00 mol Fe2O3*(-1652 kJ/2 moles of Fe2O3)=???
****Answer should contain three significant figures.
You can answer the third question.
For your first question, you have to determine the limiting reagent:
10.0g Fe*(1 mole/55.845g)= moles of Fe
2.00g O2*(1 mole/32.00g)= moles of O2
Eyeballing it, I think O2 is the limiting reagent.
Proceed as you would for the second problem.
Then you don't understand the question because that equation tells you everything you ever wanted to know and then some.
Here is what it says. It tells you that if 4 mols Fe reacts with 3 mols O2 it will produce 2 mols Fe2O3 and 1652 kJ of heat.
So the first question is 4 mols Fe and xs O2, how much heat? What is your answer to that?