amount of zinc required to produce 1.12 ml of hydrogen at STP on treatment with dilute HCl
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Answered by
36
heya....
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
tysm..#gozmit
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
tysm..#gozmit
Answered by
5
Zn + 2HCl -----> H2 + ZnCl2
Now it is given in the question that 1.12 ml at STP is formed
look at the equation carefully
you will see
1 mole of Zn gives you 1 mole of H2
so ...yea its on this basis we will be solving the rest of this question
1.12 ml of H2 is how many moles
1.12/22.4 = .05 moles
so 0.05 moles of Zn is required
0.05 * 65.4 = 3.27g (by the way 65.4g is the molar mass of Zn.,,,
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