Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: arul on November 07, 2013, 02:03:24 PM
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Barium carbonate + nitric acid = barium nitrate + carbon di oxide + water.
Barium carbonate is in solid form but nitric acid is in liquid form.
what is the formula for equimolar reaction?
I know this is simple calculation but i am confused.
some people suggested to calculate using density, mass , vol formula by avoding purity nitric acid others suggested to use purity. dont know which is correct ?
Barium carbonate = M.w=197.34 g/mol
nitric acid = M.w= 63.01, density= 1.41 kg , purity = 69 %
solid + liquid ----> equimolar formula = what?
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Ignoring the question posted for now, what is your understanding of an "equimolar formula"?
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Ignoring the question posted for now, what is your understanding of an "equimolar formula"?
I guess I have not asked the question properly.
"Equimolar" represents equal number of moles in the given amount of the solution.
For solid compounds
Eg. Barium carbonate ( M.w= 197.34g/mol)
If I dissolve 197.34 grams in 1000 ml I will get 1 mole.
But for liquid compounds
Eg. Nitric acid ( M.w=63.01, density= 1.51, purity=69% )
Here I cannot take nitric acid in grams because nitric acid is in liquid form. In these case how should I take 1 mole of nitric acid in 1000 ml.
What is the correct formula for this?
If I want to mix one mole of barium carbonate and one mole of nitric acid in 100 ml solution how many grams of barium carbonate and how many ml of nitric acid I should take? how to calculate this.
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How many moles of nitric acid do you need?
What mass is it?
In what mass of solution is it?
What is volume of this solution?
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@arul
Are you starting with pure nitric acid or a aqueous solution of nitric acid of a designated concentration?
From the way you ask the question it is not clear.
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@arul
Are you starting with pure nitric acid or a aqueous solution of nitric acid of a designated concentration?
From the way you ask the question it is not clear.
The Nitric acid which I have is 69% of purity. In 100 ml of distilled water how many ml of nitric acid should I add to get 1 mole of nitric acid aqueous soution.( Here concentration is not important for me)
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@arul
Are you starting with pure nitric acid or a aqueous solution of nitric acid of a designated concentration?
From the way you ask the question it is not clear.
The Nitric acid which I have is 69% of purity. In 100 ml of distilled water how many ml of nitric acid should I add to get 1 mole of nitric acid aqueous soution.( Here concentration is not important for me)
It is not common practice to measure the number of mols in 1ml of pure substance.
You provide 69% nitric acid, is this weight / volume or volume / volume?
Normally acid concentration is given as weight per unit volume i.e. the % is grams per 100 millilitres.
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@arul
Are you starting with pure nitric acid or a aqueous solution of nitric acid of a designated concentration?
From the way you ask the question it is not clear.
The Nitric acid which I have is 69% of purity. In 100 ml of distilled water how many ml of nitric acid should I add to get 1 mole of nitric acid aqueous soution.( Here concentration is not important for me)
It is not common practice to measure the number of mols in 1ml of pure substance.
You provide 69% nitric acid, is this weight / volume or volume / volume?
Normally acid concentration is given as weight per unit volume i.e. the % is grams per 100 millilitres.
Yes 69 % represents weight/volume.
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Great so every 1ml contains 0.69g of nitric acid.
How many mols of nitric acid in 0.69g?
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Great so every 1ml contains 0.69g of nitric acid.
How many mols of nitric acid in 0.69g?
moles = weight of the substance taken/ molecular weight
moles of nitric acid = 0.69/63.01 = 0.01095.
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The Nitric acid which I have is 69% of purity. In 100 ml of distilled water how many ml of nitric acid should I add to get 1 mole of nitric acid aqueous soution.( Here concentration is not important for me)
You should now be able to calculate this.
You haven't answered my original question
Ignoring the question posted for now, what is your understanding of an "equimolar formula"?
It is still not clear what you have been asked to calculate / report from your experiment.
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The Nitric acid which I have is 69% of purity. In 100 ml of distilled water how many ml of nitric acid should I add to get 1 mole of nitric acid aqueous soution.( Here concentration is not important for me)
You should now be able to calculate this.
You haven't answered my original question
Ignoring the question posted for now, what is your understanding of an "equimolar formula"?
It is still not clear what you have been asked to calculate / report from your experiment.
Thanks much
My understanding for "equimolar formula" is equal mole calculation for two compounds ( A & B ). If two compounds are solid I know the method but for solid(A) and liquid (B) combination Is there any "formula ".
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Can you write a balanced equation for the reaction:
Barium carbonate + nitric acid :rarrow: barium nitrate + carbon dioxide + water.
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Can you write a balanced equation for the reaction:
Barium carbonate + nitric acid :rarrow: barium nitrate + carbon dioxide + water.
I didnt access my email for the last two days. Sorry for late reply
2HNO3 + BaCO3 ----------->Ba(NO3)2 + H2O + CO2
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Ok so if your equation is correct
2HNO3 + BaCO3 :rarrow: Ba(NO3)2 +CO2 + H2O
Then this is not an equimolar reaction, you have two mols of nitric acid and only one mole of Barium carbonate.
So I am a bit confused by what you are asking, I apologise if I am being a bit slow on the uptake.
My understanding for "equimolar formula" is equal mole calculation for two compounds ( A & B ). If two compounds are solid I know the method but for solid(A) and liquid (B) combination Is there any "formula ".
Let us assume that both are solids, replace any volumes in ml to grams. How would you solve this problem?
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Ok so if your equation is correct
2HNO3 + BaCO3 :rarrow: Ba(NO3)2 +CO2 + H2O
Then this is not an equimolar reaction, you have two mols of nitric acid and only one mole of Barium carbonate.
So I am a bit confused by what you are asking, I apologise if I am being a bit slow on the uptake.
My understanding for "equimolar formula" is equal mole calculation for two compounds ( A & B ). If two compounds are solid I know the method but for solid(A) and liquid (B) combination Is there any "formula ".
The above mentioned equation is not an example of equimolar ratio (1:1). It is an example of (2:1) ratio. 2 moles of nitric acid and 1 mole of barium carbonate.
I thought if i can find out how to calculate one mole of nitric acid I can find out 2 moles by myself.
Let us assume that both are solids, replace any volumes in ml to grams. How would you solve this problem?
For two solid components . for Example
Compound "A" --> glycine ( M.w = 75.07 g/mol )
If i dissolve 75.07 grams in 1000 ml i get one mole.
so in 100 ml solution if I dissolve 7.5 grams I will get 1 mole.
similarly for compound "B"
Barium nitrate --> ( M.W = 261.337 g/mol)
If i dissolve 261.337 grams in 1000 ml I get one mole.
so in 100 ml solution if I dissolve 26.13 grams I will get 1 mole.
So in a 100 ml solution If I take 7.5 g grams of glycine and 26.1 grams of barium nitrate this is called equimolar ratio ( 1:1) . This is what I am following for solid + solid combination. Correct me If I am wrong?
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If i dissolve 75.07 grams in 1000 ml i get one mole.
What is the molar concentration then?
so in 100 ml solution if I dissolve 7.5 grams I will get 1 mole.
No, you don't. You get 0.1 mole.
If i dissolve 261.337 grams in 1000 ml I get one mole.
What is the molar concentration then?
so in 100 ml solution if I dissolve 26.13 grams I will get 1 mole.[/quote[
No, you don't. You get 0.1 mole.
Still, 0.1:0.1 means equimolar.
Do you see now the link between concentrations and moles?
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If i dissolve 75.07 grams in 1000 ml i get one mole.
What is the molar concentration then?
Molar concentration is 1 M
so in 100 ml solution if I dissolve 7.5 grams I will get 1 mole.
No, you don't. You get 0.1 mole.
If i dissolve 261.337 grams in 1000 ml I get one mole.
What is the molar concentration then?
Molar concentration is 1 M
so in 100 ml solution if I dissolve 26.13 grams I will get 1 mole.[/quote[
No, you don't. You get 0.1 mole.
Still, 0.1:0.1 means equimolar.
Do you see now the link between concentrations and moles?
My apology I miscalculated 1 mol instead of 0.1 mol for 100 ml solution. thanks for correcting me.
Yes Now I see the relation between moles and concentration. If I want 2M solution of barium nitrate Instead of 26.13 grams in 100 ml if I dissolve 52.26 grams in 100 ml i will get. Till here I am clear with how to take equimolar ratio of two solid substances and to make 1M, 2M concentrations.
But how to take equimolar ration of any Liquid+ liquid compounds as well as solid + liquid compounds.
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Equimolar means identical number of moles. Doesn't matter if substances come in the form of a solution (in which case number of moles can be easily calculated by solving molar concentration definition [itex]C=\frac n V[/itex] for n), solid or gas.