The amount of zinc required to produce 1.12ml of H2 at STP on treatment with dilute HCL will be ?
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Answered by
357
The balanced equation is :
Zn + 2HCl ---> ZnCl₂ + H₂
1mole of Zn produces 1 mole of H₂
1mole of Hydrogen gasoccupies 22.4lit ( 0r 22400ml)at STP.
22400ml of Hydrogen gas is produced from 1mole of Zn(65g)
22400ml of H₂ ------> 65g of Zn
1.12ml of H₂ -----?
=( 65*1.12)/22400
= 3.25 x 10⁻³g
=32.5x 10⁻⁴g
∴AMOUNT OF Zn required is
32.5x 10⁻⁴g
Zn + 2HCl ---> ZnCl₂ + H₂
1mole of Zn produces 1 mole of H₂
1mole of Hydrogen gasoccupies 22.4lit ( 0r 22400ml)at STP.
22400ml of Hydrogen gas is produced from 1mole of Zn(65g)
22400ml of H₂ ------> 65g of Zn
1.12ml of H₂ -----?
=( 65*1.12)/22400
= 3.25 x 10⁻³g
=32.5x 10⁻⁴g
∴AMOUNT OF Zn required is
32.5x 10⁻⁴g
Answered by
3
Answer is 32.5×10*-4
Hope this will be helpful for you.
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