[sundrops]

H3C6H5O7(aq) + NaHCO3(s)  CO2(g) + H2O(l) + Na3C6H5O7(aq)


A student measures 13.38 g of baking soda and adds this to 22. mL of 4.0 M isocitric acid (in an alcoholic solution).

Identify the limiting reagent and calculate the number of moles of CO2 produced.

so, 13.38g NaCO3/84.008g/mol NaCO3 = 0.15927molNaCO3
0.15927molNaCO3 * 3mol citric acid/1molNaC)3 = 0.478mol citric acid

and, 0.022L * 4.0M = 0.088 mol citric acid
0.088mol citric acid * 1molNaCO3/3mol citric acid = 0.0293 mol NaCO3

therefore the limiting reagent is baking soda.

right?
and i'm not too sure how to calculate the amount of CO2 produced.

any ideas?

[mike]

First step is to balance the equation.

Once this is done you determijne the limiting reagent (as you have already done).

Once you know the limiting reagent you can use that to determine how much of each product would form if the reaction went to completion (100%).

For example the ratio of bicarbonate (reactant) to CO2 (product) is 1:1 so if 0.159mol of bicarbonate react this will form 0.159mol of CO2, from this you can work out the mass of CO2 produced.

[sundrops]

H3C6H5O7(aq) + 3NaHCO3(s)  -> 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)

kay the equation is balanced and it is still a 1:1 ratio
and there is: 0.159 mol baking soda so there is also 0.159 moles of CO2!

yes!  
thanks!

[sundrops]

as a continuation to the above problem:
hypothetically if the temperature of the solution goes from 26degrees celcius to 8 degrees celcius, then calculate the temperature change per mole of H+ which reacted.

so the formula would be:

change in temperature / mol H+ reacted

so -18 degrees celcius + 273.15 degrees kelvin / mol H+ reacted = ?? K/mol

now i tried to find the moles H+ the same way we found moles of CO2 - but my answer was wrong :S

so to keep up with the board rules, i'm gonna post my work anyways,
3 moles of H+ react for every citric acid
so; 0.088mol citric acid *3 mol H+ = 0.264
255.15/.264 = 966.47 K/mol

but thats not correct - can anyone point me in the right direction?

[mike]

Ok, so first the change in temperature is 18C which is the same as 18K (remember that is is change in temperature you are calculating).

Second, remember you calculated the number of moles of the limiting reagent. You said the limiting reagent is bicarb. correct. So the number of moles of the limiting reagent is 0.159mol.

For every 3 moles of bicarb, 1 mole of citric acid will react, so for 0.159 moles of bicarb, 0.053 moles of citric acid will react (note that this means that there will be some citric acid left over after all the bicarb has reacted, hence why it is the limiting reagent.).

So now you know the number of moles of citric acid, n, (0.053mol) and the change in temperature 18C(or K). Remeber there a three hydrogens per citric acid molecule.

Hope this helps.

Mike

[AWK]

 

H3C6H5O7(aq) + 3NaHCO3(s)  -> 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)

The problem of this reaction is far more complex
PKa of H2CO3 - 6.4 and 10.3
pKa of isocitric acid - 3.29, 4.71and 6.40
From this data the stoichiometry 1:3 is rather wrong.
Fortunately, there is something more equivalents of isocitric acid  for stoichiometry 1:2 than needed and all NaHCO3 can be converted to H2CO3 (=CO2(aq) )
moles CO2 = moles NaHCO3

[mike]

hmmm I am not sure exactly why it is "far more complex" why do you think the stoichiometry is wrong?

What do you think the answer is then?

[AWK]

Since strong acid replaces the weaker one (taking into account pKa values, H3C6H5O7(aq) and NaH2C6H5O7(aq) react with NaHCO3 completely, not Na2HC6H5O7(aq) )only this reaction:
H3C6H5O7(aq) + 2NaHCO3(s)  -> 2CO2(g) + 2H2O(l) + Na2HC6H5O7(aq)
can proceeds completely.
the next step is an equlibrium:
Na2HC6H5O7(aq) + NaHCO3(s)  <-> CO2(g) + H2O(l) + Na3C6H5O7(aq)

moles of NaHCO3 - 0.1593
moles of isocitric acid 0.088

For the first reaction we need 0.0796 moles of isocitric acid, its excess is 0.00836 mole, but since ther is no NaHCO3 at this moment, the first reaction proceeded completely and 0.1593 mole of CO3 is were produced

Though calculations of Sundrops are wrong,  the answer concerning limiting reagent, surprisingly,  is correct, and of course moles of CO2 calculated by Mike are correct.