Hydrogen gas is prepared in the laboratory by reacting dilute - HCl- with granulated zinc- Following reaction takes place- Zn - 2HCl - ZnCl-2 - H-2-What would be the volume of hydrogen gas is liberated at STP when 32-65 g of zinc reacts with - HCl- 10-03 L- 11-57 L- 11-35 L- 9-53 L-

The correct option is C 11.35 L Balanced reaction is: Zn+2HCl → ZnCl_2+H_2 Molar mass of Zn= 65.4 g/molthus 65.4 g of zinc liberates 1 mol i.e. ...

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Standard XII

Chemistry

Preparation of Hydrogen

Hydrogen gas ...

Question

Hydrogen gas is prepared in the laboratory by reacting dilute $H C l$ with granulated zinc. Following reaction takes place:
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$
What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with $H C l$ ?

10.03 L

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11.35 L

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11.57 L

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9.53 L

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Solution

The correct option is C $11.35 L$
Balanced reaction is:
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$
Molar mass of $Z n =$ 65.4 g/mol
thus 65.4 g of zinc liberates 1 mol i.e. 22.4 L hydrogen gas at STP
32.65 g of zinc will liberate: $\frac{22.4}{65.4} \times 32.65 = 11.18 L$ of hydrogen gas at STP
Hence, option $B$ is correct.

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Similar questions

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place

$Z n + 2 H C l \to Z n C l_{2} + H_{2}$

Calculate the volume of hydrogen gas liberated at STP when 32.65 g of zinc reacts with HCl. 1 mol of a gas occupies 22.7 L volume at STP; atomic mass of Zn = 65.3u

When dilute HCl reacts with Zinc metal, the gas liberated is?

Q. Hydrogen gas is prepared in the laboratory by reacting dilute HCl

Q. What volume of hydrogen at NTP will be liberated when 3.25 g of zinc completely dissolve in dilute HCl?

[At. mass of Zn = 65]

Q. Assertion :Zinc liberates hydrogen gas when reacted with dilute hydrochloric acid, whereas copper does not. Reason: Zinc is above hydrogen in reactivity series.

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Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction takes place:Zn+2HCl→ZnCl2+H2What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with HCl?

The correct option is C 11.35 LBalanced reaction is:Zn+2HCl→ZnCl2+H2Molar mass of Zn= 65.4 g/molthus 65.4 g of zinc liberates 1 mol i.e. 22.4 L hydrogen gas at STP32.65 g of zinc will liberate: 22.465.4×32.65=11.18L of hydrogen gas at STPHence, option B is correct.

Hydrogen gas is prepared in the laboratory by reacting dilute $H C l$ with granulated zinc. Following reaction takes place:
$Z n + 2 H C l \to Z n C l_{2} + H_{2}$
What would be the volume of hydrogen gas is liberated at STP when 32.65 g of zinc reacts with $H C l$ ?