10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded volume of gaseous product at STP after reaction is :

22.4

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2 \times 22.4

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3 \times 224

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4 \times 22.4

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Solution

The correct option is A $22.4$ L
$2 H_{2} + O_{2} ⟶ 2 H_{2} O$

i.e, $4 g$ of $H_{2}$ reacts with $32 g$ of $O_{2}$

Now, $64 g$ of $O_{2}$ reacts with $8 g$ of $H_{2}$

$∴ H_{2}$ remaining= $10 - 8 = 2 g$

$∴$ Volume of $2 g$ of $H_{2} = 22.4 L$

Hence the correct option is A.

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10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded volume of gaseous product at STP after reaction is :

The correct option is A 22.4 L2H2+O2⟶2H2Oi.e, 4g of H2 reacts with 32g of O2Now, 64g of O2 reacts with 8g of H2∴H2 remaining= 10−8=2g∴ Volume of 2g of H2=22.4LHence the correct option is A.

The correct option is A $22.4$ L
$2 H_{2} + O_{2} ⟶ 2 H_{2} O$

i.e, $4 g$ of $H_{2}$ reacts with $32 g$ of $O_{2}$

Now, $64 g$ of $O_{2}$ reacts with $8 g$ of $H_{2}$

$∴ H_{2}$ remaining= $10 - 8 = 2 g$

$∴$ Volume of $2 g$ of $H_{2} = 22.4 L$

Hence the correct option is A.