KClO3 in a 0.1862 g sample of an explosive was determined by the reaction with 50.00 mL of Fe2+ 0.01162 M. ClO3- + 6 Fe2+ + 6H+ → Cl- + 3H2O + 6Fe3+ When the reaction was complete, the excess Fe2+ was back-titrated with 13.26 mL of Ce4+ 0.07654 M. Calculate the percentage of KClO3 in the sample.

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Q: determined by titrating it with a 0.1342 M Ce4+ solution. The balanced net ionic equation for the…

A: Given-> Molarity of Ce^4+ = 0.1342M Volume of Ce^4+ = 19.33 ml Volume of H3AsO3 = 25.00ml

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Q: 12.47 mL of a 0.33 M KMnO4 was used

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Q: 11. A suspension of milk of magnesia, Mg(OH)2, is analysed by back titration. A 1.0391 g sample is…

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Q: A student is given 1.1110 g of pure CuO. To recover the Cu present in the compound, the dark powdery…

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