10 grams of a sample of potassium chlorate on complete combustion gave 2.24l of oxygen at ntp
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no of moles of KClO4 =10/138.
no of moles at stp=2.24/2.24=1mol.
percentage=(no of moles of kclo4 /no of moles at stp ).100=(10/138).100=7.24%
ANS:7.24%
no of moles at stp=2.24/2.24=1mol.
percentage=(no of moles of kclo4 /no of moles at stp ).100=(10/138).100=7.24%
ANS:7.24%
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