What volume of oxygen is required to burn completely $90 d m^{3}$ of butane under similar conditions of temperature and pressure?
$2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$

585 d m^{3}

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90 d m^{3}

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180 d m^{3}

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435 d m^{3}

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Solution

The correct option is B $585 d m^{3}$
$2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$

2 volume of butane requires 13 volumeof oxygen.
So, for 90 vol butane, oxygen requires is $13 \times \frac{90}{2} = 585 d m^{3}$

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Similar questions

Q. The volume of oxygen required to burn completely 90 dm

^{3}

of butane under similar conditions of temperature and pressure is ______.

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

Q. Find the amount of oxygen required for the complete combustion of 4 mol of butane.

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

(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.

Q. Find the amount of oxygen required for the complete combustion of 4 mol of butane.

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

Q. Butane gas burns in air according to the following reaction,

2 C_{4} H_{10} + 13 O_{2} ⟶ 10 H_{2} O + 8 {C O}_{2}

Suppose the initial and final temperatures are equal and high enough so that all reactants and products act as perfect gases. Two moles of butane are mixed with

13

moles of oxygen and then completely reached. Find the final pressure (if the volume remains unchanged and the pressure before the reaction is

P_{0}

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What volume of oxygen is required to burn completely 90dm3 of butane under similar conditions of temperature and pressure?2C4H10+13O2→8CO2+10H2O

The correct option is B 585dm32C4H10+13O2→8CO2+10H2O2 volume of butane requires 13 volumeof oxygen.So, for 90 vol butane, oxygen requires is 13×902=585 dm3

What volume of oxygen is required to burn completely $90 d m^{3}$ of butane under similar conditions of temperature and pressure?
$2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$

The correct option is B $585 d m^{3}$
$2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$

2 volume of butane requires 13 volumeof oxygen.
So, for 90 vol butane, oxygen requires is $13 \times \frac{90}{2} = 585 d m^{3}$