[csavage]

We did an experiment in chemistry class and now I am stuck on the end questions.

The chemical reaction in regards to the experiment is:

NaHCO3 + HCl --> H2O + CO2 + NaCl

The question is:

Calculate the predicted yield of NaCl (in grams) assuming your unknown is sodium hydrogen carbonate.

My data:

Mass of empty beaker: 49.459 g

Mass of beaker and unknown: 49.687 g

Mass of unknown: 0.228 g

Moles of unknown assuming it is sodium hydrogen carbonate: 0.00271406 moles

Mass of beaker and sodium chloride: 49.940 g

Mass of sodium chloride: 0.481 g

Moles of sodium chloride: 0.00823927 moles

My Results

I determined that the ratio is 1:1.  So,

0.228 * 1 mol/ 84 g/mol = 0.0027 M

Molecular weight of NaCl is 58 so,

0.0027 * 58 = 0.1574

 

This is where I got stuck.

I did 0.481/0.157 *100 = 305.73 %

Can someone please tell  me where I am going wrong and correct me?

Please and thank you !

[Dan]

Your calculations are fine given the numbers you have. Whether the numbers are correct is another question.

[csavage]

Your calculations are fine given the numbers you have. Whether the numbers are correct is another question.

Thanks! Also, how would I go about determining the identity of the solution given this information?

[CafeYen]

I agree with your theoretical yield
0.228 g / 84 g (MM of NaHCO3) is 0.00271 mol NaHCO3
with 1:1 ratio ( i also double checked your rxn is balanced i got so many question wrong bc of this lol)
your theoretical yield for NaCl is 0.00271 mol * 58.5 g/mol (MM for NaCl) = 0.1585 g.
Therefore with the weight of your product at 0.481 gram, you have (0.481g / 0.1585 g)*100 = 303 % yield !
Something went wrong in your purification of the product that you probably didn't dry the crystal to the point where there's no more water in the product that's why your yield in so high. Keep that product in the dessicator and weigh it again