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Inorganic Compunds

  • Thread starter Adam2987
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  • #1
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Homework Statement


1.a, How many grams of sulphuric acid will neutralize 10.0g of sodium hydroxide?
b, What volume of water vapour at 100 degrees Celcius and 110 kPa would also be produced?


Homework Equations





The Attempt at a Solution


a) my answer for a is 12.3 g, didn't have issues with this question

b) I know I'm suppose to use PV=nRT I just can't figure out how to go about the question, do I calculate the number of moles for H2O first? Or do I use one of the moles from the previous calculations? Don't need the full answer just a push in the right direction.
 

Answers and Replies

  • #2
Borek
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Can't say I understand your problem.

How much water is produced in the reaction?
 
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  • #3
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Let me show you my work.

for a)

H2SO4 + 2NaOH ----> Na2SO4 + 2H2O

NaOH H2SO4 molar ratio 2:1
2 mol 1 mol

Molar mass NaOH:
Na: 1 x 23.0 g = 23.0g
OH: 1x 17.0g = 17.0g
= 40.0g

moles of NaOH reacted: (10.0g)/(40.0g/mol) = 0.250 or 2.50x10^-1 mol

Let x = mol of H2SO4 x/1 = 0.25/2 = 0.125
0.250 mol x mol
2NaOH --> H2SO4
2 mol 1 mol

Molar Mass H2SO4 = 98.1g
Thus, mass H2SO4 produced:
= 0.125 x 98.1g/mol
= 12.3g

Therefore 12.3 g of sulphuric acid will neutralize 10.0 g of sodium hydroxide

b) What volume of water vapour at 100 degrees Celcius and 110 kPa would also be produced?

Here's what I did:

H2SO4 + 2NaOH ----> Na2SO4 + 2H2O
PV=nRT, therefore, V = P/nRT : V = 110kPa/(0.250mol)(8.31kPa x L / mol x K) (373K)

V= 0.142L
Therefore 0.142 L of water vapour would be produced.

Now if this is correct, I'm not sure why. A friend of mine told me to focus on the NaOH mol as it was directly involved in the question. Not sure what it meant, so if this is correct, an explantion would be great :D

Also, any explaination on fractional coeffiecients would be great, I've read pages and pages about it, but I'm not sure how I can produce them when creating compounds, in my opinion it's easier to just use coefficients but the chapter in my text books says fractional coefficients are alot easier in compund design and understanding, I just can't seem to understand it's simplicity if there is one.
 
  • #4
Borek
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28,398
2,800
Let me show you my work.
You dodged my question, at least partially.

moles of NaOH reacted: (10.0g)/(40.0g/mol) = 0.250 or 2.50x10^-1 mol
Correct.

x/1 = 0.25/2 = 0.125

Thus, mass H2SO4 produced:
= 0.125 x 98.1g/mol
= 12.3g
Correct.

Here's what I did:

H2SO4 + 2NaOH ----> Na2SO4 + 2H2O
PV=nRT, therefore, V = P/nRT : V = 110kPa/(0.250mol)(8.31kPa x L / mol x K) (373K)
You have not shown where did the 0.25 mole of water came from - that's what I was asking about. But you are right - for each mole of NaOH neutralized there is one mole of H2O produced, so if you started with 0.25 mole of naOH you have produced 0.25 mole of water - that's the amount of substance that will be now treated as a gas.

But you are using wrong formula, so the result is off. PV=nRT so V is NOT P/nRT.

Also, any explaination on fractional coeffiecients would be great
No idea what you mean - please give an example.
 
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  • #5
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Me and you have the same question, the question as stated is all the information I've been given, in my original post I asked if I was suppose to calculate the moles of water on my own or use a previous molar quantity, a friend told me to use the mol of NaOH with no explaination. Not trying to doge your question, I just can't answer it.

as for my fractional coeffiecient issues, here's an example

Al + 3/2 H2SO4 ---> 1/2 Al2(SO4)3 + 3/2 H2


If I were to solve the first part, how do I go about getting 1/2 Al2(SO4)3 + 3/2 H2
I fully understand balancing equations using coefficients, but this fractional coefficient stuff just makes my brain hurt. I just don't understand how you go about getting 1/2 an Al2 ion.
 
  • #6
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If I'm not suppose to use PV=nRT to calculate the volume of water vapour produced, would I use a mass-volume calculation? I'm unsure because this ust really seemed like a PV=nRT situation
 
  • #7
Borek
Mentor
28,398
2,800
Me and you have the same question, the question as stated is all the information I've been given, in my original post I asked if I was suppose to calculate the moles of water on my own or use a previous molar quantity, a friend told me to use the mol of NaOH with no explaination.
Do you know how to read balanced reaction equation? Stoichiometric coeffcients tell you what is ratio of moles of ALL substances involved, doesn't matter if they are reactants or products. To calculate amount of water you use exactly the same approach you used to calculate amount of sulfuric acid. Look at the reaction equation, look at the stoichiometric coeffcients, use them for calculations.

Not trying to doge your question, I just can't answer it.
Yes you can.

as for my fractional coeffiecient issues, here's an example

Al + 3/2 H2SO4 ---> 1/2 Al2(SO4)3 + 3/2 H2


If I were to solve the first part, how do I go about getting 1/2 Al2(SO4)3 + 3/2 H2
I fully understand balancing equations using coefficients, but this fractional coefficient stuff just makes my brain hurt. I just don't understand how you go about getting 1/2 an Al2 ion.
You don't get half of an atom, but once you have the equation balanced in fractional coefficients you can always convert it easily to whole numbers. So don't worry too much.

If you try to balance

H2 + O2 -> H2O

you can assume there was one molecule of oxygen - then it is obvious there are two molecules of water produced and that means two molecules of hydrogen were necessary. However, if you start assuming there was one molecule of hydrogen it follows that you produced one molecule of water - but that in turn means only half a molecule of oxygen was used:

H2 + 1/2O2 -> H2O

Just multiply by two to get rid of fraction and everything is OK.

Note that in this case it was more or less obvious that starting with oxygen will not yield fractional coefficients. In the case of more complicated reaction equations it is not always that obvious and sometimes you have balanced everything else and you are left with the last element - and you can either throw away everything already done and start again, or use fractions to balance - and get rid of them later. Guess which approach is faster.
 
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  • #8
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Thank you for the help, much appreciated. Fractional coefficients make more sense now haha. I understand what you're saying about how to go about calculating the water vapour, but when I reflect on my text book if I go about the way I got H2SO4, It would have to be at STP no? But because we are given a temperature and a pressure, we have to use PV=nRT
Atleast thats what my text says. I go about it mathematically, I still get the same answer. The book also makes reference to PV = nRT therefore V = P/nRT. So combining what you say and what the text says, I'm even more confused than I was before. The only difference now is that I know why we make reference to NaOH, which clears up alot.

Thank again for your help, much appreciated.
 
  • #9
Borek
Mentor
28,398
2,800
PV=nRT works for gases, not for everything. At 100 deg C water is gaseous, hence you can use ideal gas equation to calculate its volume. Sulfuric acid is liquid at these temperatures and its volume is not that easy to calculate.

The book also makes reference to PV = nRT therefore V = P/nRT.
Check your math, you got it reversed.
 
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  • #10
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oh... @!$## V = nRT/P so my answer would change drastically, gives me 7.05 L
haha whoa thanks for pointing that out
 

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