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T = 133 + 273 = 406 K

R= 0.0821 L.atm/Kmol

V = 126 L

P= 0.880 atm

n (number of moles) = M/molar mass of KClO3= g/122.54

PV= nRT

PV= M/122.54 RT

M = 407.6 g

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Q: What mass of KClO3 must be decomposed to produce 126 L of oxygen gas at 133 degrees C and 0.880 ATM?
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