0.23 gram na is treated with dilute sulphuric acid to give h2 which is converted into water by reacting with oxygen which come from how many moles of potassium chlorate on heating

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Vikas TU · Answer

Dear Student Na + H2SO4 => Na2SO4 + H2 2H2 + O2 => 2H2O 2KClO 3 ------> 2KCl+3O 2 2 mole on heating gives 3 mle of O2 And we require 1 mole of 02 to react with H2 and give water . So, we need 2/3 mole of KclO3 .

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0.23 gram na is treated with dilute sulphuric acid to give h2 which is converted into water by reacting with oxygen which come from how many moles of potassium chlorate on heating

0.23 gram na is treated with dilute sulphuric acid to give h2 which is converted into water by reacting with oxygen which come from how many moles of potassium chlorate on heating

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0.23 gram na is treated with dilute sulphuric acid to give h2 which is converted into water by reacting with oxygen which come from how many moles of potassium chlorate on heating

0.23 gram na is treated with dilute sulphuric acid to give h2 which is converted into water by reacting with oxygen which come from how many moles of potassium chlorate on heating

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