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Given following reaction, NaClO3 + Fe → O2 + FeO + NaCl

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Given following reaction, 

NaClO3 + Fe → O2 + FeO + NaCl 

In the above reaction 492 L of O2 is obtained at 1 atm & 300 K temperature. Determine mass of NaClO3 required (in kg). 

(R = 0.082 L atm mol–1 K–1 ) 

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1 Answer

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by (50.2k points)

mol of NaClO3 = mol of O2

mol of O2 = PV/RT = 1 x 492/0.082 x 300 = 20 mol

mass of NaClO3 = 20 × 106.5 = 2130 g

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