Grams of gas equation reacted question.?

Question

When heated, solid KClO3 forms solid KCl and O2 gas. A sample of KClO3 is heated and 226 mL of gas with a pressure of 749 mm Hg is collected over water, at 26°C. At 26°C, the vapor pressure of water is 25 mm Hg.

2KClO3(s) → 2KCl(s) + 3O2(g)

How many grams of KClO3 were reacted?

Answer 1

You're going to need the Ideal Gas Law:
PV = nRT
but you're going to have to do some pruning to your knowns first.

Current conditions
T = 26 deg C = 299K
V = 226 mL = 0.226 L
P = 749 mm Hg - 25 mm Hg (partial pressure of H2O vapor needs to subtracted to get the P of the actual O2 gas.)
= 724 mm Hg = 0.953 atm (div. by 760 to change to atm)
R = 0.0821 lit-atm/mole-K

PV = nRT
(0.953 atm)(0.226L) = (moles)(0.0821)(299K)
0.00877 moles O2

0.00877 moles O2 are generated. But how many moles
of KClO3 were needed to make that much O2? (Pssst.
STOICHIOMETRY)

Per your balanced equation, 3 moles O2 requires
2 moles KClO3, so 0.00877 moles O2 needs
0.00585 moles KClO3

0.00585 moles KClO3 * 122.6 g / mole = 0.717 g KClO3

Answer 2

Moles of KClO3 = (2/3)Moles of Oxygen

= (2/3)(724*.226/.0821*299)

Answer 3

28G of beans = copious hydrogen sulfides and methane gasses . ha ha .