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solmaris
1 month ago
6

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 pre

ssure 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?
Chemistry
1 answer:
KiRa [2.8K]1 month ago
6 0

Answer:

moles = 0.093 moles

Explanation:

Here, we analyze a reaction occurring in a vessel with a total pressure of 730 torr.

The total pressure can be expressed as:

Pt = Pwater + PO2

We need to ascertain the pressure of O2 since it will allow us to find the moles of KClO3 through stoichiometry.

The oxygen pressure is:

PO2 = 730 - 26 = 704 Torr

Next, we need to convert this pressure from Torr to Atm:

704 Torr divided by 760 Torr equals 0.9263 atm

Now, using the ideal gas equation:

PV = nRT

We can calculate the moles of O2, and consequently the moles of KClO3:

n = PV/RT

R equals 0.082 L atm /K mol

P equals 0.9263 atm

V equals 3.72 L

T equals 27 + 273 equals 300 K

Plugging in the values:

n = 0.9263 * 3.72 / 300 * 0.082

n = 0.14 moles

Finally, using stoichiometry, since 2 moles of KClO3 yield 3 moles of O2, we have:

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

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Here's my calculation

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