the laboratory instructor gives you a test tube Laboratory Questions Pre- show A

Question

the laboratory instructor gives you a test tube Laboratory Questions Pre- show ALL WORK TO RECEIVE CREDIT. A dean, porcelain crucible weig and the mass is recorded. A mixture containing and KCI is added to the crucible and the crucible is reweighed. NaH Mass of crucible and lid 18.498 g Mass of crucible, lid, and mixture 19734 g The crucible and mixture are heated for 15 minutes, then allowed to cool to room temperature. The crucible and residue are weighed and the mass is recorded. Mass of crucible, lid, and residue after 1" heating 19.476 g The crucible and residue are reheated for 5 minutes, then allowed to cool to room temperature. The crucible and residue are weighed and the mass is recorded. Mass of crucible, lid, and residue after 2nd reheating 19.368 g The crucible and residue are reheated for 5 minutes, then allowed to cool to room temperature. The crucible and residue are weighed and the mass is recorded. Mass of crucible, lid, and residue after 3rd reheating 19.369 g 1. Determine the mass of the mixture used in the experiment.

Explanation / Answer

Q1.

mass of mixture in experiment:

Mass of mixture = (Mass of Cruicible + lid + mixture ) - (Mass of Cruicible + lid) = 19.734-18.498 = 1.236 g

Q2.

ratio for mass loss of CO2: H2O of gams

mass lost = 19.734 - 19.369 = 0.365 g lost

this must be CO2 and H2O

mass of NaHCO3 in mix:

44g of CO2 + 18g of H2O

fraction of H2O = 18/(18+44) = 0.290

fraction of CO2 = 44/(18+44) = 0.710

mass of H2O --> 0.365 *0.29 = 0.10585 g

mass of CO2 --> 0.365 *0.71 = 0.25915 g

Q3.

total mass of CO2 and H2O after 3 cycles:

Mass lost = Mass of cruicible + lid + mix - ( Mass of cruicible + lid) after third try = 19.734 - 19.369 = 0.365 g

Q4.

from:

mass of H2O --> 0.365 *0.29 = 0.10585 g

mass of CO2 --> 0.365 *0.71 = 0.25915 g

calculate

mol of CO2 = mass/MW = 0.25915 /44 = 0.005889

mol of H2O = mass/MW = 0.10585 /18 = 0.00588

note that each, 1 mol of CO2 and 1 mol of H2O should come from 1 mol of NaHCO3

so

0.00588 mol of NaHCO3 were present

mass = mol*MW = 0.00588*84.007 = 0.49396 g of NaHCO3

Q5

% of NaHCO3 and KCl --> mass of NaHCO3 / total * 100% = 0.49396 / 1.236 *100 = 39.96%

so

KCl --> (100-39.96)% = 60.04%

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