[autopsy]

For our experiment we used potassium bicarbonate, and 6 M hydrochloric acid.
KHCO3 + HCL ---> KCl + H2O + CO2

so Potassium bicarb and HCl react with one another at a 1:1 mole ratio. Also every molke of potassium bicarb present will create  1 mole of potassium chloride will be created.
So we wighed 2.5 grams of potassium bicarb in a dish weighed it, than dissolved it in 5mL water. we than added 6mL of 6 M HCl. We than put this dish on top of a beaker that was our water bath and heated it on a bunsen burner.  Once all of the liquid was gone, we reweighed it as few times, until it stopped decreasing. My starting weight was 2.30 g, my end weight was 2.14 g. 
The question in the post lab asks, Why is the mass of KCL recovered less than the starting mass of KHCO3? Theoretically, Why should the moles of KHCO3 and the moles of KCl produced be the same?
If you guys could help me with these questions, I would greatly appreciate it.         
              Steph

[AWK]

Moles are the same, but weights are different. Should be about 75 % of the weight of KHCO3. Moreover 6 mL of 6M HCl is not enough to convert all KHCO3 to KCl. Hence you have mixture of KCl and K2CO3 since KHCO3 will decompose during warming at least partially.

[autopsy]

Thank you Awk! Suppose however, i used 3.00 grams of K2CO3, insteado f the KHCO3....
The balanced equation would it be
 K2CO3 + 2 HCl --> 2KCl + H2O + CO2

And if that is correct. How many of mL of 6.0 M Hcl would I need to convert all of it?
And How many grams of KCl would be formed in the reaction?


Can you please show me the calculations, so I can learn how you came to them. Than when I am tested onhem I can know exactly what to do.

   Thanks!!!
           Steph

[Borek]

Moreover 6 mL of 6M HCl is not enough to convert all KHCO3 to KCl.

2.5 g of KHCO₃ is 0.025 mole, 6 mL of 6M HCl is 0.036 mole. 44% excess. Perhaps you calculated for K₂CO₃? (although even then it could be enough, as carbonate is for sure not anhydrous).

[Borek]

Can you please show me the calculations, so I can learn how you came to them. Than when I am tested onhem I can know exactly what to do.

Not exactly your question, but in a way all stoichiometric calculations are alike:

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

The only additional thing is how to calculate number of moles of substance in the solution - C=n/V is your starting point.

[autopsy]

Thank you for the link. But it is sort of confusing me because their examples are all single replacement and mine is not. Can anyone just show me step by step how to set it up. I have never seen or used the formula - C=n/v so far in class, so that is confusing as well. I dont have to be spoon fed the answers, but at the same time I am not good with teaching myself from various websites because the examples havent even matched up so I cant follow along.

[Borek]

Thank you for the link. But it is sort of confusing me because their examples are all single replacement and mine is not.

Type of reaction doesn't matter - when doing stoichiometric calculations only thing that matters is balanced chemical reaction equation. Once you have that you know all about stoichiometry. If one mole reacts with one mole it doesn't matter whether is it synthesis or something else.

I have never seen or used the formula - C=n/v so far in class, so that is confusing as well.

If you are told that you are using 6M solution for sure you know what molar concentration is - and C=n/V is just a definition. C is concentration, n is number of moles, V is volume. You know volume, you know concentration - solve for the number of moles.

For potassium carbonate reaction - how many moles of potassium carbonate is 3 grams? With how many moles of HCl will it react (it is in stoichiometric coefficients in the balanced reaction equation)? What volume of 6M solution contains that number of moles?

[AWK]

Moles are the same, but weights are different. Should be about 75 % of the weight of KHCO3. Moreover 6 mL of 6M HCl is not enough to convert all KHCO3 to KCl. Hence you have mixture of KCl and K2CO3 since KHCO3 will decompose during warming at least partially.

Mz Fault. Though i told about KHCO3 i thought about K2CO3 since something was wrong in this problem. Then in fact carbonate was a limiting reagent