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The amount of Zinc required to produce 1.12 ml of H2 at STP on treatment with dilute HCL will be? plz ansr soon Thanks in advance

The amount of Zinc required to produce 1.12 ml of H2 at STP on treatment with dilute HCL will be? plz ansr soon Thanks in advance

Grade:11

3 Answers

Beenu Mathew
533 Points
7 years ago
HI,
Zn + 2HCl -----> H2 + ZnCl2
Now it is given in the question that 1.12 ml at STP is formed
so????
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.,,,,checked it on google ;) )
THANKS!!
BEST OF LUCK 
CHEERS!!
 
Akash Biswas
30 Points
6 years ago
Zn + 2HCl --» ZnCl2 + H21 Mole of zinc reacts to produce 1 mole of hydrogen gas ..So 1.12 ml of hydrogen gas contains 1.12 ml/22400 ml = 0.5* 10 ^ -4Thus 0.5* 10 ^ -4 mole if hydrogen gas is obtained from 0.5* 10 ^ -4 mole of Zinc So amount of zinc required to produce 1.12 ml of hydrogen gas is Molecular weight of Zn*0.5* 10 ^ -4=65.38*0.5* 10 ^ -4= 32.5 * 10 ^ -4 g
Yashvi
15 Points
4 years ago
Zn + 2HCl= ZnCl2 + H2(g)
1 mol(65 g - atomic weight) Zn required for 1 mol H2(22.4l=22400ml)
If 65g Zn - 22400ml
Then x g Zn - 1.12 ml
1.12×65/22400= 0.00325 i.e. 3.25×10-4
 

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