How many grams of KO2 would be required to produce 1120.0L of O2 at 20.0 degrees Celsius and 1.00atm?

Answer 1

Use.

PV = nRT

To get moles O2. ( 20.0o C = 293.15 Kelvin )

(1.00 atm)(1120.0 L) = (X moles)(0.08206 L*atm/mol*K)(293.15 K)

Moles = 1120.0/24.56

= 45.60 moles O2

-----------------------------------now,

45.60 moles O2 (1 mole KO2/2 mole O2)(71.1 grams/1 mole KO2)

= 1621 grams potassium oxide required

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