Potassium chlorate on heating decomposes as under:
$2 K C l O_{3} △ - \to 2 K C l + 3 O_{2}$
Calculate (i) volume of oxygen evolved at S.T.P. (ii) weight of potassium chloride formed, when $4.9 g$ of potassium chlorate decomposes completely.
$[K = 39; O = 16; C l = 35.5]$ .

Open in App

Solution

(i) $O_{2} = 1.344 d m^{3}$
(ii) $K C l = 2.89 g$ .

Suggest Corrections

Similar questions

Q. Potassium chlorate decomposes, on heating, to form potassium chloride and oxygen. When 24.5 g of potassium chlorate is decomposed completely, then 14.9 g of potassium chloride is formed. Calculate the mass of oxygen formed. Which law of chemical combination have you used in solving this problem?

Potassium nitrate on strong heating decomposes as under:
2KNO₃ $\overset{∆}{\to}$ 2KNO₂+ O₂
Calculate:
(i) weight of potassium nitrite formed.
(ii) weight of oxygen formed when 5.05 g of potassium nitrate decomposes completely.
(K = 39, O = 16, N = 14)

Q. Calculate the mass of potassium chloride formed and the volume of oxygen evolved at 27C and 750 mm of mercury, when 24.5 g of potassium chlorate is heated.

Q. Calculate the mass of oxygen obtained by complete decomposition of 10 kg of pure potassium chlorate (Atomic mass K = 39, O = 16 and Cl= 35.5)

Q. When potassium chlorate

(K C l O_{3})

is heated in the presence of manganese dioxide catalyst, it decomposes to form potassium chloride and oxygen gas. Represent this in the form of a balanced chemical equation.

Potassium chlorate on heating decomposes as under:2KClO3△−→2KCl+3O2Calculate (i) volume of oxygen evolved at S.T.P. (ii) weight of potassium chloride formed, when 4.9 g of potassium chlorate decomposes completely.[K=39;O=16;Cl=35.5].

(i) O2=1.344 dm3(ii) KCl=2.89 g.

Potassium chlorate on heating decomposes as under:
$2 K C l O_{3} △ - \to 2 K C l + 3 O_{2}$
Calculate (i) volume of oxygen evolved at S.T.P. (ii) weight of potassium chloride formed, when $4.9 g$ of potassium chlorate decomposes completely.
$[K = 39; O = 16; C l = 35.5]$ .

(i) $O_{2} = 1.344 d m^{3}$
(ii) $K C l = 2.89 g$ .