`49.0g` of an impure sample of `KClO_(3)` on heating gives `6.72` lit of oxygen gas at STP. The percentage of impurity of `KClO_(3)` is [Given that `KClO_(3)rarr KCl+O_(2)`; unbalanced]

x' g of KCIO_3 on decomposition gives 'y' ml of O_2 at STP. The percentage purity of KCIO_3 would be :

500 g impure sample of CaCO_(3) on heating gives 70 g of CaO . Percentage impurities in sample is

33.6 g of an impure sample of sodium bicarbonate was heated strongly and it gave 4.4 g CO_2 . The percentage purity of NaHCO_3 will be:

245 g impure sample of KClO_3 on heating gives 12gO_2(g) according to 2KClO_3(s)rarr2KCl(s)+3 O_2(g) Calculate % purity of sample ?

100 cm^(3) of a given sample of H_(2)O_(2) , gives 1000cm^(3) of O_(2) at S.T.P. The given sample is

A given initial mass of KClO_(3) on 50% decompostion produces 67.2 litre oxygen gas at 0^(@)C and 1 atm. The other product of decomposition is KCl. The initial mass of KClO_(3) (in gm) taken is:

An impure sample ( having 60 % purity ) of KClO_(3) contains 30 gm of pure KClO_(3) .Find weight of impure sample.

The decomposition of KClO_(3) to KCl and O_(2) on heating is an example of :

KClO_(3) on heating decomposes to give KCl and O_(2) . What is the volume of O_(2) at N.T.P liberated by 0.1 mole of KClO_(3) ?

Balance the following equation : KClO_(3) to KCl + O_(2)