Question

Oxygen gas, generated by the reaction 2KClO3(s)---2KCl(s)+3O2(g), is collected over water at 27•C in 3.72L vassel at a total pressure of 730 torr.(The vapor pressure of H2O at 27•C is 26.0 torr). How many moles of KClO3 were consumed in the reaction?
Oxygen gas, generated by the reaction 2KClO3(s)---2KCl(s)+3O2(g), is collected over water at 27•C in a 1.16L 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?

Answer 1

Answer:

moles = 0.093 moles

Explanation:

In this case, we know that this reaction is taking plave in a vessel that has a 730 torr of total pressure.

The total pressure is a value obtained by:

Pt = Pwater + PO2

We need to know the pressure of O2, because then, with stoichiometry, we can calculate the moles of KClO3

The pressure of oxygen is:

PO2 = 730 - 26 = 704 Torr

Now, this pressure is in Torr, and we need to convert it to Atm, so:

704 Torr / 760 Torr = 0.9263 atm

Now, let's use the ideal gas equation:

PV = nRT

With this expression, we will calculate the moles of O2, and then, the moles of KClO3:

n = PV/RT

R = 0.082 L atm /K mol

P = 0.9263 atm

V = 3.72 L

T = 27 + 273 = 300 K

Replacing the data:

n = 0.9263 * 3.72 / 300 * 0.082

n = 0.14 moles

Finally, by stoichiometry, we know that 2 moles of KClO3 produces 3 moles of O2, so:

moles of KClO3 = 0.14 * 2/3 = 0.093 moles of KClO3