Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: jamesbrown on April 03, 2016, 10:32:11 AM
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If the specific heat capacity of water is 4.2 J/g/°C and the mass of 1cm3 of water is 1g
Then In an experiment, 0.40g of butane (C4H10) was completely burned in air. The heat was given off raised the temperature of 100g of water by 54.2°C.
So calculate the heat released by burning 0.40g of fuel in KJ.
Would the answer be 0.40*54.2*4.2 = 91.056 KJ
or did I do something wrong? because it said put it in KJ but I didn't account for that I don't think?
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If the specific heat capacity of water is 4.2 J/g/°C and the mass of 1cm3 of water is 1g
Then In an experiment, 0.40g of butane (C4H10) was completely burned in air. The heat was given off raised the temperature of 100g of water by 54.2°C.
So calculate the heat released by burning 0.40g of fuel in KJ.
Would the answer be 0.40*54.2*4.2 = 91.056 KJ
or did I do something wrong? because it said put it in KJ but I didn't account for that I don't think?
If you had less water to heat, would you expect the temperature rise to be more, less, or the same?
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Less?
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Not less, no. Consider the energy as fixed, like e.g. having a pound of rice to share amongst first 8, then 4 people. Each person gets more in the second case.
You haven't taken into account the mass of water in your calculation.
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What is the equation that I am suppose to use then I will be able to work it out!
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If the specific heat capacity of water is 4.2 J/g/°C and the mass of 1cm3 of water is 1g
Then In an experiment, 0.40g of butane (C4H10) was completely burned in air. The heat was given off raised the temperature of 100g of water by 54.2°C.
So calculate the heat released by burning 0.40g of fuel in KJ.
Would the answer be 0.40*54.2*4.2 = 91.056 KJ
or did I do something wrong? because it said put it in KJ but I didn't account for that I don't think?
Hint: Why do you need to measure the temperature of water?
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You need to measure the temperature to get the temperature rise when the reaction has completed then you can put it in the equation q = mcΔt
so m is the mass of a substance in which the heat is being transferred.
so c is the specific heat capacity (which is 4.2 in this case)
so Δt is the temperature rise
and q is the heat energy given out/released
so q = 100*4.2*54.2
so q = 22764 KJ
Is this right? I think I was getting confused with the 0.4g of fuel as the next part is calculating moles!
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You need to measure the temperature to get the temperature rise when the reaction has completed then you can put it in the equation q = mcΔt
so m is the mass of a substance in which the heat is being transferred.
so c is the specific heat capacity (which is 4.2 in this case)
so Δt is the temperature rise
and q is the heat energy given out/released
so q = 100*4.2*54.2
so q = 22764 KJ
Is this right? I think I was getting confused with the 0.4g of fuel as the next part is calculating moles!
You get the mass right now. But there's another problem. The specific heat capacity(c) that you are using is 4.2 J/g/°C, note the unit. So, why are you using kJ as your unit in the calculation? what shall you do so as to use the unit kJ as required in the question?
Once you handled this, you can go to the next part.
Edit: I overlooked the original question that told us to use kJ as the energy unit.
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So are you saying I need to change the specific heat capacity to KJ rather than J and then work it out again so:
4.2J = 0.0042KJ
So then it would be q = 100*0.0042*54.2
so q = 22.764 KJ
is that right or was my answer before right?
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So are you saying I need to change the specific heat capacity to KJ rather than J and then work it out again so:
4.2J = 0.0042KJ
So then it would be q = 100*0.0042*54.2
so q = 22.764 KJ
is that right or was my answer before right?
right.
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So question 1 was:
Calculate the heat released by burning 0.40g of fuel in KJ
And we found that the answer was 22.764KJ
The question 2 is:
Calculate the heat released in KJ/g of fuel burned
Could someone explain what to do for this one?
Thanks!
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The question 2 is:
Calculate the heat released in KJ/g of fuel burned
When you burn 0.4g of the fuel you get 22.764KJ
So what amount of energy will you get when you burn 1g of the fuel?
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so would you get 56.91KJ of energy produced?
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That's what I got
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then for question 3 it says calculate the moles of fuel burned:
which is 0.4/58
moles = 0.0069
I think
then question 4 says calculate the heat released in KJ/mol of fuel burned
so for that would it be 22.76/0.0069 or is that wrong?
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*Ignore me, I am impatient*
Could someone please check this for me as I think its wrong.
Thankyou
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then for question 3 it says calculate the moles of fuel burned:
which is 0.4/58
moles = 0.0069
I think
then question 4 says calculate the heat released in KJ/mol of fuel burned
so for that would it be 22.76/0.0069 or is that wrong?
Correct.
*Ignore me, I am impatient*
Please read the forum rules. (http://www.chemicalforums.com/index.php?topic=65859.0). Don't bum p. Be patient.
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Thanks you!
Also another thing say If the specific heat capacity of water is 4.2 J/g/°C and the mass of 1cm3 of water is 1g
then 100cm3 of 0.20 mol/dm3 copper sulphate solution was put in a calorimeter and 2.0g of magnesium power added. The temperature of the solution rose by 25.1 °C.
Calculate the heat released in KJ.
So would you do q = m*Δt*c
q = 0.2*25.1*0.0042
q = 0.021KJ
Is this right because the concentration part is confusing me?
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What is mass of the solution?
Concentration is there just to let you find out the limiting reagent in the Mg/CuSO4 reaction.
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Do I have to add an extra 100g to the mass because 1cm3 is 1g and it state sthere is 100cm3
So the calculation would be 100.2*25.1*0.0042
Is this right, I'm kind of stuck at he moment due to our teacher not giving us usable notes
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Do I have to add an extra 100g to the mass because 1cm3 is 1g and it state sthere is 100cm3
So the calculation would be 100.2*25.1*0.0042
Is this right, I'm kind of stuck at he moment due to our teacher not giving us usable notes
For the equation Q=mcΔT
The m refers to the mass, but not molarity or concentration, of the substance.
Your 100.2 seems to come from 100g of solution + 0.2 mol dm-3 of solution, which is not correct.
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So what is the mass then or could you tell me how to find it? I don't really get it
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Sorry is the mass 102 then? not 100.2 I was reading the wrong thing
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Sorry, but I start getting a little confused.
then 100cm3 of 0.20 mol/dm3 copper sulphate solution was put in a calorimeter and 2.0g of magnesium power added. The temperature of the solution rose by 25.1 °C.
Number of moles of magnesium= 2/24.3 = 0.0823 mol
Number of moles of copper sulphate= 0.1*0.2= 0.02 mol
Magnesium is in excess. Not all magnesium will be dissolved. Also, some copper metal is formed too. Calculating the mass in this case can be complex, is it (100g + mass of magnesium unreacted + mass of copper formed)? Please someone answer this.
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Exactly, its hard for people like you and I just started being taught it like a week ago. So I think you can see where I am coming from and why I need help.
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You have 100 mL of solution. Let's assume its density is that of water (they differ a bit, but the difference is not large). Do you remember what is the density of water? If you know volume, and density, can you calculate mass of the solution?
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All I know is the specific heat capacity of water is 4.2 J/g/°C and the mass of 1cm3 of water is 1g
100cm3 of 0.20 mol/dm3 copper sulphate solution was put in a calorimeter and 2.0g of magnesium power added. The temperature of the solution rose by 25.1 °C.
And I was told to calculate the heat released in the reaction in KJ
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If you know volume, and density, can you calculate mass of the solution?
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Sorry thats all I was told about are you sure theres no way of doing this?
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There is a very simple way of calculating mas of the solution, which comes directly from the density definition. You are expected to know such things.
Please note that you should never think in terms of a "it is not a part of what we have learned yesterday". Every piece of everything you have learned in the past matters.
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Oh sorry man I thought you were saying that you couldn't do it because there wasn't enough information. You should have been more clear.
so density=mass/volume
so mass = volume*density
the volume is 100
the density is 0.2
so the mass = 20
is this correct?
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Oh sorry man I thought you were saying that you couldn't do it because there wasn't enough information. You should have been more clear.
My question was perfectly clear from the very beginning.
the density is 0.2
No, 0.2 is not the density.
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So are we trying to find the density or the mass? if its the mass could you please help me out and tell me what the density is ?
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then 100cm3 of 0.20 mol/dm3 copper sulphate solution was put in a calorimeter and 2.0g of magnesium power added. The temperature of the solution rose by 25.1 °C.
Number of moles of magnesium= 2/24.3 = 0.0823 mol
Number of moles of copper sulphate= 0.1*0.2= 0.02 mol
Magnesium is in excess.
If you know volume, and density, can you calculate mass of the solution?
@Borek
So we can just calculate the mass of the solution, neglecting the mass of undissolved magnesium and formed copper?
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So are we trying to find the density or the mass? if its the mass could you please help me out and tell me what the density is ?
We are trying to find the mass, and you correctly stated what the density of the solution is at least twice in your earlier posts.
By definition, density is a mass of 1 mL of the substance. What is the mass of 1 mL of water?
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So we can just calculate the mass of the solution, neglecting the mass of undissolved magnesium and formed copper?
To be precise we should take into account everything we put into the calorimeter (and you are perfectly right that magnesium mass needs to be taken into account), I am just trying to guide jamesbrown through the first step of the process.
Note that you are making the problem harder than it is. Because of mass conservation it doesn't matter what reactions took place, mass of the mixture after the reaction is exactly the same as it was before the reaction.
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What is the mass of 1 mL of water?
Is it 100g or 0.1kg?
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What is the mass of 1 mL of water?
Is it 100g or 0.1kg?
Neither. And 100g and 0.1 kg is exactly the same number.
the mass of 1cm3 of water is 1g
And why do you guess, if you have already posted a correct number?
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Okay then btw I did know they were the same I just wasn't sure what unit you wanted it in?
Whta the next step then?
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So we can just calculate the mass of the solution, neglecting the mass of undissolved magnesium and formed copper?
To be precise we should take into account everything we put into the calorimeter (and you are perfectly right that magnesium mass needs to be taken into account), I am just trying to guide jamesbrown through the first step of the process.
Note that you are making the problem harder than it is. Because of mass conservation it doesn't matter what reactions took place, mass of the mixture after the reaction is exactly the same as it was before the reaction.
Thanks.
Okay then btw I did know they were the same I just wasn't sure what unit you wanted it in?
Whta the next step then?
You know that 1mL of water is 1g. So what is the mass of the 100mL solution(assume the solution's density is same as water)?
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1ml of water is 1g so 100ml of the solution would have to be 100g
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Ok, you have mass of solution to be 100g. Now, what is the mass of the solids(magnesium and copper) formed? With these two numbers you can find out the total mass.
Hint:
Because of mass conservation it doesn't matter what reactions took place, mass of the mixture after the reaction is exactly the same as it was before the reaction.
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Well it says 2g of magnesium was added. and im not sure but I think 100g was copper?
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100g is the mass of the solution not copper.
Borek have said that mass of the mixture after the reaction is exactly the same as it was before the reaction, so you only need to calculate the initial mass of the mixture, which is the mass of magnesium added and the mass of the solution. (Sorry for confusing you again with the mass of copper).
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Its all good so the overall initial mass is 102g right?
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Yes
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So the equation is q=102*0.0042*25.1
so q = 10.75..
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Yes
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So the next question is calculate the moles of copper sulphate reacting. So the mr is 160 and the mass is 100g so would it be 0.625 moles?
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100cm3 of 0.20 mol/dm3 copper sulphate solution was put in a calorimeter and 2.0g of magnesium power added. The temperature of the solution rose by 25.1 °C.
(For convenience I quote the numbers given in here again)
So the next question is calculate the moles of copper sulphate reacting. So the mr is 160 and the mass is 100g so would it be 0.625 moles?
No. 100g is the mass of the solution which contains water and copper sulphate. You can't directly calculate the number of moles from this.
This problem involves the concept of limiting reagent. Start by calculating the number of moles of both copper sulphate and magnesium first. For a solution, the number of moles is given by (volume)*(molarity).
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Why do I need to calculate the moles of magnesium because its not even part of the solution wouldn't it be better to calculate the moles of water and then the solution and then subtract. Also what do you mean by morality, is it another word for concentration?
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Why do I need to calculate the moles of magnesium because its not even part of the solution
Because you need to know which is the limiting reagent for this reaction(have you learnt that?)
wouldn't it be better to calculate the moles of water and then the solution and then subtract.
How?
Also what do you mean by morality, is it another word for concentration?
Yes.
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No we dont know what a limiting reagent is sorry?
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Oh... start reading from this: https://en.wikipedia.org/wiki/Limiting_reagent
And, write a balanced equation for the reaction first.
By the way, is this a high school homework or what? Your school didn't teach limiting reagent when teaching mole calculations?
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Its high school I don't think they did i can't remember. Could you link an easier tutorial Wikipedia is way too complex, or could you just explain it yourself?
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https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/limiting-reagent-stoichiometry/v/stoichiometry-limiting-reagent
It's a fairly basic concept. Let's say you have a cake recipe that uses 5 cups of flour and 1 cup of sugar. You have 10 cups of flour in your pantry and 3 cups of flour. Is flour or sugar going to be your "limiting reagent" in this case?
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Did you mean to say 10 cups of flour in your pantry and 3 cups of sugar or is what you said correct?
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Yes. Sorry.
Don't worry too much about the analogy. That Khan Academy video should give you a good understanding of limiting reagents.
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I found out that the moles of copper sulphate was 20, because the volume is 100 and the density is 0.2 so there is 20 moles.
The the next question was what is the moles of magnesium, which I worked out to be 0.083, because ar = 24 and mass = 2.
But the next question was show that magnesium was in excess? can someone explain how to do this? is it empathy change.
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What are the units of density? Not sure where you're finding the value of 0.2 for it either?
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I found out that the moles of copper sulphate was 20, because the volume is 100 and the density is 0.2 so there is 20 moles.
No. You have a 0.2M copper sulphate solution of 100 cm3. The molarity is 0.2M(or mol dm-3) and the volume is 100cm3, i.e. 0.1 dm3. To calculate the number of moles in a solution, you need to multiply the molarity(concentration) by the volume, but not density. And bear in mind the units, you need to convert cm3 to dm3 before the calculation(1 dm3 = 1000 cm3)
The the next question was what is the moles of magnesium, which I worked out to be 0.083, because ar = 24 and mass = 2.
Good
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So is the moles of copper sulphate 0.2*0.1=0.02 moles, or is this wrong and can you also help me with this next question:
the next question was show that magnesium was in excess? can someone explain how to do this? is it empathy change.
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So is the moles of copper sulphate 0.2*0.1=0.02 moles
Yes
the next question was show that magnesium was in excess? can someone explain how to do this? is it empathy change.
No, not 'empathy' or enthalpy or such kind of things.
So you have number of moles of Mg=0.083 mol and number of moles of CuSO4=0.02 mol. The reactant in excess is the one which will not be used up when the other reactant(s) is used up. To determine this, you need to write a balanced equation of the reaction between CuSO4 and Mg first.
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The balanced equation is: 2CuSO4 + 2Mg :rarrow: 2MgSo4 + Cu2
So whats the next step?
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The balanced equation is: 2CuSO4 + 2Mg :rarrow: 2MgSo4 + Cu2
Incorrect. Why do you write Cu2? Cu exists as metals, not molecules.
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2CuSO4 + 2Mg :rarrow: 2MgSo4 + 2Cu
is this right, then tell me what the next step is?
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is this right, then tell me what the next step is?
You were told several times this is a limiting reagent problem at this stage, and you were told where to look for hints and explanations. Instead of trying to find an information about how to solve this kind of a problem, you only ask "what is the next step, what is the next step, what is the next step". Sorry, this is going nowhere, you are not going to learn anything this way. I am locking this thread, let it be a lesson for you - nobody is going to spoon feed you for ever.