xperiment was decomposing 2KCIO3 (s) Unknown #3 Experiment: Mass of test tube 38

Question

xperiment was decomposing 2KCIO3 (s) Unknown #3 Experiment: Mass of test tube 38.548 g Mass of KClO3 1.133 g Mass of KC1O3 and test tube 39.681 g Mass of KCIO3 and test tube after heat 39.411 g of produced 220 mL Volume Need help with calculation K! Data is above, thank you! Calculated Results for Unknown t3 Experiment: A) Mass of On produced 0.27 g B) Theoretical mass of that should have been produced if the yield was 0.00924 g 100% 0.0139 moles c) Theoretical moles of O based on mass 0.00992 moles d) Moles of K Cio3 required 1.1325 e) Mass of KCIO3 required for CO2 produced f Percentage of KCIO3 in unknown mixture based on mass of C2 produced 99.96% 0.343 L g) Liters of that theoretically should have been produced h) Theoretical moles of O2 that should have been produced ifyield was 100% 0.0139 mol 2.94% Moles of KC13 required 1.70g j) Mass of KClO3 required for O2 Percentage of KClo in unknown mixture based on volume of O: roduced

Explanation / Answer

The question seems to have data missing. I am not quite getting the source from where you got all the volume and mass results. Please clarify. I will work out the problem with the understanding that you are asking for what percentage of the theoretical requirement of KClO3 you actually started with. Is that right?

1) Mass of test tube = 38.548 g

2) Mass of KClO3 = 1.133 g

3) Mass of test tube + KClO3 = 39.681 g

4) Mass of KClO3 and test tube after heat = 39.411 g

5) Volume of O2 produced = 220.0 mL

g) Liters of O2 that theoretically should have been produced = 0.343 L

h) Theoretical moles of O2 that should have been produced if yield was 100% = (0.343 L)*(1 mole/22.4 L) (molar volume = volume of 1 mole of gas at STP = 22.4 L) = 0.0153 mol.

i) Moles of KClO3 required = Look at the balanced equation above. The molar ratio of KClO3 and O2 is 2:3.

Therefore, moles of KClO3 = (0.0153 mol O2)*(2 mole KClO3/3 mole O2) = 0.0102 mol (ans).

j) Mass of KClO3 required for O2 =

Molar mass of KClO3 = 122.55 g/mol.

Therefore, mass of KClO3 required = (0.0102 mol)*(122.55 g/1 mol) = 1.25 g.

k) Percentage of KClO3 based on volume of O2 = You required 1.25 g KClO3 for the experiment theoretically, but you started with 1.133 g KClO3. Therefore, percentage of KClO3 = (1.133/1.25)*100 = 90.64% (ans)

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