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 :

134.4 L

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89.6 L

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22.4 L

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112 L

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Solution

The correct option is A
134.4 L

C $_{4}$ H $_{8}$ + 6O $_{2}$ $\to$ 4CO $_{2}$ + 4H $_{2}$ O
22.4 lit89.6 lit. 72 g
At S.T.P. at S.T.P.
1 mole 4 mole $\frac{72}{18} = 4$ mole
For complete combustion of 1 mole C $_{4}$ H $_{8}$
6 mole O $_{2}$ required
$n_{O_{2}} = 6 m o l e$
$V_{O_{2}} = 6 \times 22.4$
$V_{O_{2}} = 134.4 l i t .$

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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 :

Q. When

22.4 L

C_{4} H_{8}

S T P

is burnt completely,

89.6 L

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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

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When 22.4 L of C4H8 at STP is burnt completely, 89.6 L of CO2 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 :

The correct option is A 134.4 L C4H8 + 6O2 → 4CO2 + 4H2O 22.4 lit89.6 lit. 72 g At S.T.P. at S.T.P. 1 mole 4 mole 7218=4mole For complete combustion of 1 mole C4H8 6 mole O2 required nO2=6mole VO2=6×22.4 VO2=134.4lit.

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 :