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Hydrogen gas is prepared in the laborato...

Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc, Following reaction takes place
`Zn+2HClrarrZnCl_(2)+H_(2)`
Calculate the voluem 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

Text Solution

Verified by Experts

Given that, mass of Zn=32.65g
1 mole of gas occupies =22.7L volume at STP
Atomic mass of Zn=65.3u
The given equation is
`Zn+2HClrarrZnCl_(2)+H_(2)`
`65.3g 1"mol"=22.7L` at STP
From the above equation it is clear that
`65.3g Zn`. when reacts with HCl produces `=22.7L "of" H_(2)` at STP
`therefore 32.65gZn`. when reacts with HCl, will produce `=(22.7xx32.65)/(65.3)=11.35L` of `H_(2)` at STP.
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Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc. Following reaction taken place : Zn+2HCl rarr ZnCl_(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 = 65.3 u.

Calculate the volume of hydrogen liberated at STP when 500 cm^(3) of sulphuric acid reacts with excess of zinc. (H = 1, O = 16, S = 32)

Knowledge Check

  • The volume of hydrogen gas liberated at STP when a current of 5.36 ampere is passed through dil. H_(2)SO_(4) for 5 hours will be

    A
    5.6L
    B
    11.2L
    C
    16.8L
    D
    22.4L

  • What volume of hydrogen at NTP will be liberated was is 3.25 g of zinc completely dissolve in dilute HCl? (At. mass of Zn=65)

    A
    1.12litre
    B
    11.20litre
    C
    2.24litre
    D
    22.40litre

  • In the reaction Zn(s) + 2H^(+)(aq.) rarr Zn^(2+)(aq.) + H_(2) (g) , how many liters of hydrogen gas measured at STP is produced when 6.54g of Zn is used (Zn = 65.4 u) ?

    A
    `22.4 L`
    B
    `11.2 L`
    C
    `2.24 L`
    D
    `1.12 L`

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