Question

Amount of zn required to produce 1.12ml Of h2 on treatment with dil hcl will be

Answer 1

                              Zn  +  2HCl  -------->  ZnCl₂  +  H₂

From the above chemical equation,
1mole of Zn is required for preparing 1mole of H₂.


22400 millitres of volume is occupied by 1mole of H₂

1.12 millitres of volume is occupied by 5 x 10⁻⁵ moles of H₂

So, 5 x 10⁻⁵ moles of Zn is required to prepare 1.12ml of H₂ gas.