Consider the reaction of zinc metal with hydrochloric acid: Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g).

If 29.39 × 1024 atoms of zinc completely reacted with hydrochloric acid, how many moles of hydrochloric acid must have reacted?

Do NOT include units in your entry. Report your answer with 3 SFs.

moles of HCl

6.054×10²⁵

Explanation:

1)find number of moles of zinc

2)multiply the mole of zinc with 2

3)use the formula mol = number of particle/ avogadro constant

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%