Experiment was decomposing 2KCIO3G 2KCle) 302G) Unknown #3 Experiment: 38.548 g

Question

Experiment was decomposing 2KCIO3G 2KCle) 302G) Unknown #3 Experiment: 38.548 g Mass of test tube Mass of KC103 1.133 g 39.681 g Mass of KClO3 and test tube Mass of KCIO3 and test tube after heat 39.411 g 220. mL. Lvolume of O2 produced Need help with calculation I! Data is above, thank you! Calculated Results for Unknown #3 Experiment: 0.27 g A) Mass of O2 produced B Theoretical mass of O2 that should have been produced if the yield was 0.00924 g 100% 0.0139 moles c) Theoretical moles of O2 based on mass 00992 moles d Moles of KClO3 required e) Mass of KClO3 required for O2 produced 99.96% 0.343 L h) Theoretical moles of O2 that should have been produced ifyield was 100%. 0.0139 mol Moles of KCla required j) Mass of KC103 required for O2 k) Percentage of KClO3 in unknown mix based on volume ofo: produced

Explanation / Answer

Ans

a)

Assume all mas "no longer in tube" is O2

so,

Oxygen mass = 39.681-39.411 = 0.27 g of O2

b)

if yield was 100% then

mol of KClO3 = mass/MW = 1.133/122.55 = 0.009245

expect, 0.009245 mol of KCl and 3/2*0.009245 = 0.0138675 mol of O2 expected

mass = mol*MW = 0.0138675*32 = 0.44376 g of O2

d)

moles of KClO3 required --> 0.009245

e)

mass of KClO3 = 0.009245*122.55 = 1.1325

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