Hi there.
There are some concerns regarding this experiment.
1. The chemical equation you have written is correct. You can further form carbonic acid from H2O and CO2. Take a look at this:
Na2CO3 + 2HCl ---> 2NaCl + H2O +CO2 eq1
H2O + CO2 ---> H2CO3 (carbonic acid) eq2
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Na2CO3 + 2HCl ---> 2NaCl + H2CO3 eq 3
You can use the equation 3 to solve for the concentration of carbonic acid.
2. The calculation for the number of moles of Na2CO3 is correct. We have concerns for the concentration of HCl. Please specify the concentration of the HCl reagent that you used. You cannot simply assume 50ml = 50 grams. Remember that you used a solution of HCL meaning that 50 mL HCl solution has concentration. Please ask your instructor what is the concentration of HCL that has been used in the experiment.
If you have the concentration already, let say it is in molarity, you can just get the number of moles by multiplication.
moles HCl = (concentration of HCL reagent) x (50 mL)
Now from here you can proceed in selecting the limiting reagent.
3. Solving for the concentration of the carbonic acid can be accomplished using the formula:
moles carbonic acid/ L solution
Moles of carbonic acid can be solve using stoichiometry. Just be sure item 2 is solved. The total volume of the solution is needed as well.
hope this helps :)
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