When $22.4 L$ of $C_{4} H_{8}$ at $S T P$ is burnt completely, $89.6 L$ of $C O_{2}$ gas the $S T P$ and $72 g$ of water are produced. The volume of the oxygen gas the STP consumed in the reaction is closest to

89.6 L

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112 L

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134.4 L

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22.4 L

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Solution

The correct option is D $134.4 L$
$C_{4} H_{8} 22.4 l i t A t S . T . P . + 6 O_{2} \to 4 C O_{2} + 4 H_{2} O 89.6 l i t . 72 g a t S . T . P .$

$1 m o l e 4 m o l e \frac{72}{18} = 4 m o l e$

For complete combustion of $1 m o l e C_{4} H_{8}$

$6 m o l e O_{2}$ required

$n_{O_{2}} = 6 m o l e$

$V_{O_{2}} = 6 \times 22.4$

$V_{O_{2}} = 134.4 l i t .$

Hence, the correct option is $C$

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Similar questions

When 22.4 L of C $_{4}$ H $_{8}$ at STP is burnt completely, 89.6 L of CO $_{2}$ gas at STP and 72 g of water are produced. The volume of the oxygen gas at STP consumed in the reaction is closest to :

29 g

of butane is burnt in excess of

O_{2}

at STP. Find the volume of the

C O_{2}

gas produced at STP.

Q. Which has maximum volume at

S T P

? (

1

mole of a gas

= 22.4 L

S T P

)

Q. Amount of gas which occupies

22.4 L

volume at STP is equal to vapor density of the gas.

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When 22.4 L of C4H8 at STP is burnt completely, 89.6 L of CO2 gas the STP and 72 g of water are produced. The volume of the oxygen gas the STP consumed in the reaction is closest to

The correct option is D 134.4 LC4H822.4 litAtS.T.P.+6O2→4CO2+4H2O89.6 lit. 72 gatS.T.P.1 mole 4 mole 7218=4 moleFor complete combustion of 1 moleC4H86 mole O2 requirednO2=6 moleVO2=6×22.4VO2=134.4 lit.Hence, the correct option is C

When $22.4 L$ of $C_{4} H_{8}$ at $S T P$ is burnt completely, $89.6 L$ of $C O_{2}$ gas the $S T P$ and $72 g$ of water are produced. The volume of the oxygen gas the STP consumed in the reaction is closest to