If 1 mL of Hydrogen gas reacts with 1 mL of chlorine gas at STP, the volume of hydrogen chloride gas produced at STP is:

1 mL

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3 mL

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4 mL

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2 mL

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Solution

The correct option is D 2 mL
As no.of moles $= \frac{V o l u m e o f c o m p o u n d}{V o l u m e a t S T P}$
$n_{H 2} = \frac{1 m l}{22.4 l}$
$n_{C l 2} = \frac{1 m l}{22.4 l}$
$H_{2} + {C l}_{2} ⇌ 2 H C l$
$1 + 1 ⇌ 2$
1 mole of $H_{2}$ and 1 mole of ${C l}_{2}$ gives 2 mole of HCl.
Thus $1 m l$ of $H_{2}$ and $1 m l$ of ${C l}_{2}$ gives 2ml of HCl (at STP).

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If 1 mL of Hydrogen gas reacts with 1 mL of chlorine gas at STP, the volume of hydrogen chloride gas produced at STP is:

The correct option is D 2 mLAs no.of moles =VolumeofcompoundVolumeatSTPnH2=1ml22.4lnCl2=1ml22.4lH2+Cl2⇌2HCl1+1⇌ 21 mole of H2 and 1 mole of Cl2 gives 2 mole of HCl.Thus 1ml of H2 and 1ml of Cl2 gives 2ml of HCl (at STP).