find the mass of the residue left when 0.5 mole of kc10( 3) (subscript) is heated strongly.
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2KClO(3)+heat gives 2 KCl+ 3 O(2)
mole of KClO(3) = 0.5 moles
using stoichiometry ,
mole of KClO(3)/2=mole of KCl/2
so, mole of KClO(3)= mole of KCl
Now,
mass of KCl = 0.5*74.5 = 37.25g
mole of KClO(3) = 0.5 moles
using stoichiometry ,
mole of KClO(3)/2=mole of KCl/2
so, mole of KClO(3)= mole of KCl
Now,
mass of KCl = 0.5*74.5 = 37.25g
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