O2 is evolved by heating KClO3 using MnO2 as a catalyst- 2KClO3 2 KCl - 3O2The number of moles of oxygen present in 6-72 litres is x-Please give your answer in the form of 10x-

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Standard XII

Physics

Pressure vs Velocity

O2 is evolved...

Question

$O_{2}$ is evolved by heating $K C l O_{3}$ using $M n O_{2}$ as a catalyst.

$2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}$

The number of moles of oxygen present in $6.72 l i t r e s$ is $x$ .

[Please give your answer in the form of $10 x$ ]

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Solution

Reaction is: $2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}$

We have to calculate the number of moles of $O_{2}$ present in 6.72 litres.

22.4 litres of $O_{2}$ has 1 mole.

6.72 litres of $O_{2}$ will have = $\frac{6.72}{22.4} = 0.3 m o l e s$

Therefore, $10 x = 0.3 \times 10$

$= 3$

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Similar questions

2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}

The number of moles of

O_{2}

in the above equation is:

Q. 1 mol of

K C l O_{3}

is heated in the presence of

M n O_{2}

. The produced oxygen is used to react with aluminium. Find the number of moles of aluminium oxide formed.

(a) (i) What volume of oxygen is required to burn completely $90 d m^{3}$ of butane under similar conditions of temperature and pressure?

$2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$

(ii) the vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP?
(iii) A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?

(b) $O_{2}$ is evolved by heating $K C l O_{3}$ using $M n O_{2}$ as a catalyst.

$2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}$

(i) Calculate the mass of $K C l O_{3}$ required to produce 6.72 litres of $O_{2}$ at STP.
[atomic masses of $K = 39, C l = 35.5, O = 16]$
(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.
(iii) Calculate the volume occupied by 0.01 mole of $C o_{2}$ at STP.

When is heated, it decomposes into . If some is added, the reaction goes much faster because

[CPMT 1971,76,80,94]

Q. 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]

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O2 is evolved by heating KClO3 using MnO2 as a catalyst. 2KClO3MnO2−−−−→2KCl+3O2The number of moles of oxygen present in 6.72 litres is x.[Please give your answer in the form of 10x]

Reaction is: 2KClO3MnO2−−−−→2KCl+3O2We have to calculate the number of moles of O2 present in 6.72 litres.22.4 litres of O2 has 1 mole.6.72 litres of O2 will have =6.7222.4=0.3 molesTherefore, 10x=0.3×10 =3

$O_{2}$ is evolved by heating $K C l O_{3}$ using $M n O_{2}$ as a catalyst.

$2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}$

The number of moles of oxygen present in $6.72 l i t r e s$ is $x$ .

[Please give your answer in the form of $10 x$ ]

Reaction is: $2 K C l O_{3} M n O_{2} - --- \to 2 K C l + 3 O_{2}$

We have to calculate the number of moles of $O_{2}$ present in 6.72 litres.

22.4 litres of $O_{2}$ has 1 mole.

6.72 litres of $O_{2}$ will have = $\frac{6.72}{22.4} = 0.3 m o l e s$

Therefore, $10 x = 0.3 \times 10$

$= 3$