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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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Answer» The answer is (02.13) 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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