Oxygen from the decomposition of KClO3 is collected in an eudiometer. If the mas

Question

Oxygen from the decomposition of KClO3 is collected in an eudiometer. If the mass of the KClO3 decomposed is 0.234 grams, how many moles of O2 were formed? [2KClO3 -> 2KCl + 3O2] In a similar experiment, 0.00307 moles of O2 is collected at 19 C and a partial pressure of 720 torr. The volume is measured to be 74.1 mL. What would this volume be at room conditions, 19C and 742 torr? -continuing with the same experimental results, calculate the indicated molar volume at room conditions, 720 torr and 19 C - what is the theoretical molar volume of the gas at 742 torr and 19 C ? -Calculate the percent error in this experiment

Explanation / Answer

a)

Moles of KClO3:

mass = 0.234 g of KCLO3

mol = mass/MW = 0.234/(122.55) = 0.0019094 mol of KClO3

1 mol of KClO3 = 3/2 = 1.5 mol of O2

0.0019094 mol --> 1.5*0.0019094 = 0.0028641 mol of O2

b)

Apply ideal gas law

P1V1/T1 = P2V2/T2

720*74.1/(19+273) = 742*V2/(19+273)

V2 = 71.90296 mL

c)

molar volume is given by

v = Total V / total mol

v = 71.90296 / 0.00307

v = 23,421.15 mL/mol

v = 23.4 L/mol

actual amount is about 22.4 L per mol

error:

E = (22.4-23.4)/(22.4)*100 = -4.464 %

negative implies this is a higher deviation, i.e. overestimation

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