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]

1.12 litre

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11.20 litre

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224 litre

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22.40 litre

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Solution

The correct option is B 1.12 litre
$Z n + H C l \to Z n C l_{2} + H_{2}$

When 65.38 grams of Zn reacts with $H C l$ it forms 22.4 L of hydrogen gas.

If 3.25 grams of Zn reacts with $H C l$ it forms $x$ L of hydrogen gas.

$X = \frac{3.25 \times 22.4}{65.38} = 1.11$

Hence, the correct option is $A$

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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

Q. 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

Q. What volume of hydrogen will be liberated at NTP by the reaction of Zn on 50 ml. dilute

H_{2} S O_{4}

of specific gravity 1.3 and having purity 40% ?

Q. The atomic weight of zinc is

65.4

. The volume of dry hydrogen at NTP liberated by treating

6.54 g

of zinc with dilute sulphuric acid would be:

Q. Excess of

H C l

is reacted with

6.5 g

of zinc. The hydrogen liberated is burnt with oxygen. The mass of water formed is (Atomic mass of

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= 65)

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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]

The correct option is B 1.12 litreZn+HCl→ZnCl2+H2When 65.38 grams of Zn reacts with HCl it forms 22.4 L of hydrogen gas.If 3.25 grams of Zn reacts with HCl it forms x L of hydrogen gas.X=3.25×22.465.38=1.11Hence, the correct option is A

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]