The weight of zinc required for the liberation of 10 g of hydrogen gas on reaction with $H_{2} S O_{4}$ is:

(Atomic mass of Zn is 65.38 a.m.u.)

655 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

163.7 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

491.25 g

No worries! We‘ve got your back. Try BYJU‘S free classes today!

326.9 g

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C 326.9 g
The chemical equation for the reaction between zinc and sulphuric acid can be written as:
$Z n + H_{2} S O_{4} \to Z n S O_{4} + H_{2}$
For liberation of 2 g of $H_{2}$ , amount of zinc required $= 65.38$
For liberation of 10 g of $H_{2}$ , amount of zinc required $= \frac{65.38}{2} \times 10 = 326.9 g$

Suggest Corrections

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. Mole of zinc required to liberate 448 ml of hydrogen gas at NTP on reaction with sulphuric acid is:

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

Z n

= 65)

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

Z n

= 65)

Join BYJU'S Learning Program

The weight of zinc required for the liberation of 10 g of hydrogen gas on reaction with H2SO4 is:(Atomic mass of Zn is 65.38 a.m.u.)

The correct option is C 326.9 gThe chemical equation for the reaction between zinc and sulphuric acid can be written as:Zn+H2SO4→ZnSO4+H2For liberation of 2 g of H2, amount of zinc required =65.38For liberation of 10 g of H2, amount of zinc required =65.382×10=326.9 g

The weight of zinc required for the liberation of 10 g of hydrogen gas on reaction with $H_{2} S O_{4}$ is:

(Atomic mass of Zn is 65.38 a.m.u.)