Oxygen gas,generated by the reaction is collected over water at 27°C in a 1.55-L vessel at a total pressure of 1.00 atm.(The vapor pressure of H₂O at 27°C is 26.0 torr. )How many moles of KClO₃ were consumed in the reaction? A)0.0608 moles B)0.0912 moles C)0.0405 moles D)0.0434 moles E)1.50 moles

Question

Quizplus · Accepted Answer

First, use Dalton's law of partial pressures to find the partial pressure of oxygen gas:

P(O2) = P(total) - P(H2O)

Convert the vapor pressure of water to atm:

P(H2O) = 26.0 torr = 0.034 atm

P(O2) = 1.00 atm - 0.034 atm = 0.966 atm

Next, use the ideal gas law to find the number of moles of oxygen:

PV = nRT

n = PV/RT

Convert Celsius to Kelvin:

T = 27°C + 273 = 300 K

n = (0.966 atm)(1.55 L)/(0.0821 L atm mol^-1 K^-1)(300 K) = 0.0608 mol

Finally, use the stoichiometry of the reaction to find the number of moles of KClO3 consumed:

2KClO3 -> 2KCl + 3O2

1 mol KClO3 produces 3 mol O2

Therefore,

0.0608 mol O2 x (1 mol KClO3/3 mol O2) = 0.0203 mol KClO3

But the reaction consumes 2 mol KClO3 for every 3 mol O2 produced, so the actual amount consumed is:

0.0203 mol KClO3 x (2 mol KClO3/3 mol O2) = 0.0405 mol KClO3

Therefore, the answer is choice C, 0.0405 moles.

Oxygen Gas,generated by the Reaction Is Collected Over Water