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Volume-Volume Relation

a i What volu...

Question

(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.

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Solution

a)
$i right parenthesis 2 C subscript 4 H subscript 10 plus 13 O subscript 2 rightwards arrow 8 C O subscript 2 plus 10 space H subscript 2 O 2 space v o l u m e space o f space b u tan e equals 90 space d m cubed 1 space v o l u m e space o f space b u tan e equals 45 space d m cubed t h e r f o r e v o l u m e space o f space O subscript 2 space r e q u i r e d equals 13 cross times 45 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 585 space d m cubed i i right parenthesis m o l u c u l a r space w e i g h t equals 2 cross times V D left parenthesis v a p o u r space d e n c i t y right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space equals 162 space g v o l u m e space o f space 16 space g space g a s s equals 22.4 space L v o l u m e space o f space 1 g space g a s s equals fraction numerator 22.4 over denominator 16 end fraction v o l u m e space o f space 24 space g space g a s s equals fraction numerator 22.4 over denominator 16 end fraction cross times 24 equals 33.6 space L i i i right parenthesis i t space w i l l space c o n t a i n space X space n o space o f space m o l e c u l e s$
b)

$2 KClO subscript 3 space space rightwards arrow with space space space space MnO subscript 2 space space space space space space on top space 2 KCl space space space space plus 3 straight O subscript 2 Molecular space mass space of space space 2 KClO subscript 3 space equals 2 space straight x space left parenthesis 39 plus 35.5 plus 3 straight x 16 right parenthesis space equals space 2 space straight x space 122.5 space equals 245 space straight g 2 KCl space equals 2 space left parenthesis 39 space plus 35.5 right parenthesis space equals 149 space straight g 3 straight O subscript 2 space equals space 3 space straight x space 22.4 space equals 67.2 space straight l 67.2 space litre space of space straight O subscript 2 space produced space by space KClO subscript 3 space equals 245 space straight g space 1 space litre space of space straight O subscript 2 space produced space by space KClO subscript 3 space equals fraction numerator 245 over denominator 67.2 end fraction 6.72 space of space straight O subscript 2 space produced space by space KClO subscript 3 space equals fraction numerator 245 space straight x space 6.72 over denominator 67.2 end fraction space space space space space space space space space space space space space space space space space space space space space space space equals space 24.5 space straight g space ii right parenthesis space space 245 space straight g space of space KClO subscript 3 space contains space equals space 3 space moles space of space oxygen space space space space 24.5 space straight g space of space KClO subscript 3 space contains space equals fraction numerator 3 space straight x 24.5 over denominator 245 end fraction space space space space equals 0.3 space mole since space comma space 1 space mole space of space oxygen space contain space equals 6.022 space straight x 10 to the power of 23 space molecules therefore comma 0.3 space mole space of space oxygen space contain space equals 6.022 space straight x 10 to the power of 23 space straight x 0.3 space space space space space space equals space 1.8066 space straight x space 10 to the power of 23 space molecules$

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Similar questions

Q. 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

Q. 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?

Q. A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. Under the same conditions of temperature and pressure, how many molecules of nitrogen gas would be present in the same vessel?

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

(a) Calculate the number of moles and number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?
(b) What is the vapour density of ethylene?
(Avogadro's number = 6 × 10²³; Atomic weight of C = 12, H = 1; Molar volume = 22.4 litres at STP)

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