2. If 5.4 g Al are reacted with excess HCI, how many moles of H2 will be produced? 2 Al + HCI → 2 AICI, + 3 H, (unbalanced]

0.3 moles of H2 will be produced.

Explanation:

           Balanced reaction: 2 Al + 6 HCl → 2 AlCl3 + 3 H2

Given Al mass = 5.4 g, we know Mr of Al is 27

Using the formula:

moles = mass / Mr

moles of Al = 5.4 / 27

moles of Al = 0.2 moles

Here we see that "2" molar mass before Al; so in this reaction, "1" molar mass will have [(0.2)/2] = 0.1 moles. He see "3" molar mass before H2, therefore moles of H2 is (3 * 0.1) = 0.3 moles.

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%