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### Topic: What's left?  (Read 15901 times)

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#### submar1ney

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##### What's left?
« on: January 20, 2008, 10:33:33 AM »
Given a beaker of dilute hydrochloric acid of which 12g is pure hydrochloric acid, and i add a crystal of calcite (calcium carbonate) of mass 18g, is it possible to find the mass of calcite left after the reaction is finished? If so, how?

Thank you

#### Alpha-Omega

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##### Re: What's left?
« Reply #1 on: January 20, 2008, 11:09:54 AM »
You are given:  12 g HCl + 18g CaCO3

1.   Write out your reaction equation:
CaCO3 + HCl → CaCl2 + CO2 + H2O

CaCO3 + 2HCl → CaCl2 + CO2 + H2

3.   Remember that species react mole to mole and NOT gram to gram.

4.   Then you use your balanced reaction equation and find the grams of calcite left upon the completion
of the reaction.  Which means UNREACTED or EXCESS calcite (calcium carbonate).

#### submar1ney

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##### Re: What's left?
« Reply #2 on: January 20, 2008, 03:03:47 PM »
You are given:  12 g HCl + 18g CaCO3

1.   Write out your reaction equation:
CaCO3 + HCl → CaCl2 + CO2 + H2O

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

3.   Remember that species react mole to mole and NOT gram to gram.

4.   Then you use your balanced reaction equation and find the grams of calcite left upon the completion
of the reaction.  Which means UNREACTED or EXCESS calcite (calcium carbonate).

I have already balanced it correctly, but don't understand where to go from your point 3. Can you elaborate please.

Cheers
« Last Edit: January 16, 2012, 03:05:50 PM by Arkcon »

#### Kryolith

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##### Re: What's left?
« Reply #3 on: January 20, 2008, 03:18:20 PM »
You know the moles of HCl used. From the reaction equation you know how many moles of calcite are needed to react with this amount. Convert your result to grams and substract it from the initial mass of calcite.

I hope the information helps you.

#### Borek

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##### Re: What's left?
« Reply #4 on: January 20, 2008, 03:31:58 PM »
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

#### submar1ney

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##### Re: What's left?
« Reply #5 on: January 20, 2008, 03:39:04 PM »
OK, so 1 mole of HCl = 36.51g so 2 moles = 73.02g...  12g/73.02g = 0.16 moles of HCl

1 mole of CaCO3=100g...   18g/100g = 0.18 moles of CaCO3

From there then, i take 0.16 moles from 0.18 moles to give 0.02 moles??? How does that convert to grams?

Cheers

#### Kryolith

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##### Re: What's left?
« Reply #6 on: January 20, 2008, 03:50:49 PM »
You have 12 g HCl How many moles are this? And with how much calcite can it react?

#### submar1ney

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##### Re: What's left?
« Reply #7 on: January 20, 2008, 03:59:12 PM »
You have 12 g HCl How many moles are this? And with how much calcite can it react?

I thought 1 mole of a compound was the relative molecular mass, in grams?

#### Borek

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##### Re: What's left?
« Reply #8 on: January 20, 2008, 04:02:32 PM »
You have 12 g HCl How many moles are this? And with how much calcite can it react?

I thought 1 mole of a compound was the relative molecular mass, in grams?

Molar mass doesn't depend on the reaction, it is characteristic of compound.
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#### Kryolith

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##### Re: What's left?
« Reply #9 on: January 20, 2008, 04:03:59 PM »
http://en.wikipedia.org/wiki/Molecular_mass

$n(HCl)=\frac{m(HCl)}{M(HCl)}$

Does that help to calculate the moles of HCl?

#### submar1ney

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##### Re: What's left?
« Reply #10 on: January 20, 2008, 04:09:27 PM »
No sorry guys, i'm not with you.

My book say to calculate moles, write the formula..   HCl...  and add the relative atomic masses. (not molecular mass as i previously quoted, my mistake). So, H=1.01 + Cl=35.5  =  36.51 grams - 1 mole.

No???

#### Kryolith

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##### Re: What's left?
« Reply #11 on: January 20, 2008, 04:14:08 PM »
M=molar mass

$M(HCl)=M(H) + M(Cl)=1.0079\frac{g}{mol}+35.4532\frac{g}{mol}=36.4611\frac{g}{mol}$

$n(HCl)=\frac{m(HCl)}{M(HCl)}=\frac{12g}{36.4611\frac{g}{mol}}=0.33 mol$

#### submar1ney

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##### Re: What's left?
« Reply #12 on: January 20, 2008, 04:24:26 PM »
M=molar mass

$M(HCl)=M(H) + M(Cl)=1.0079\frac{g}{mol}+35.4532\frac{g}{mol}=36.4611\frac{g}{mol}$

$n(HCl)=\frac{m(HCl)}{M(HCl)}=\frac{12g}{36.4611\frac{g}{mol}}=0.33 mol$

So 2 lots of HCl will be 0.16 moles...  which is what i calculated earlier in the thread :-) But where do i go from here to calculate how much calcite will be left over :-)

Thank you :-)

#### Borek

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##### Re: What's left?
« Reply #13 on: January 20, 2008, 04:26:44 PM »
Once again:

http://www.chembuddy.com/?left=balancing-stoichiometry&right=stoichiometric-calculations

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#### Kryolith

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##### Re: What's left?
« Reply #14 on: January 20, 2008, 04:30:21 PM »
But I don't believe you knew what you did before

So now you know that 0.33 mol react with 0.16 mol calcite. What you want to determine next is how many moles of calcite remain. Then determine the mass that remains. I am sure you can do that on your own now.

EDIT