mass of O2 by plant photos.ing 3L CO2 @T of 25C @1atm bag @roomT, .5L of gas @ 1atm. squeezed to 5atm, new vol? g of Ar in 5L balloon @STP?

Question

Roger the Mole · Accepted Answer

6 CO2 + 6 H2O → C6H12O6 + 6 O2

n = PV / RT = (1 atm) x (3 L) / ((0.08205746 L atm/K mol) x (25 + 273) K) = 0.12268 mol CO2
(0.12268 mol CO2) x (6 mol O2 / 6 mol CO2) x (31.99886 g O2/mol) = 4 g O2
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Supposing constant temperature:
(0.5 L) x (1 atm / 5 atm) = 0.1 L 
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(5 L) / (22.4 L/mol) x (39.948 g Ar/mol) = 9 g Ar

mass of O2 by plant photos.ing 3L CO2 @T of 25C @1atm bag @roomT, .5L of gas @ 1atm. squeezed to 5atm, new vol? g of Ar in 5L balloon @STP?

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