How many molecules of iron (II) phosphate awe produces when 3.6 grams of iron (II) hydroxide reacts with excess calcium phosphate

0.078×10²³ molecules

Explanation:

Given data:

Mass of iron(II) hydroxide = 3.6 g

Molecules of iron(II) phosphate = ?

Solution:

Chemical equation:

3Fe(OH)₂ + Ca₃(PO₄)₂      →     Fe₃(PO₄)₂ + 3Ca(OH)₂

Number of moles of Fe(OH)₂:

Number of moles = mass/ molar mass

Number of moles = 3.6 g/ 89.86 g/mol

Number of moles = 0.04 mol

Now we will compare the moles of iron(II) hydroxide with Fe₃(PO₄)₂:

            Fe(OH)₂            :              Fe₃(PO₄)₂

                3                   :                    1

               0.04              :                1/3×0.04 = 0.013

Number of molecules of  Fe₃(PO₄)₂:

1 mole = 6.022×10²³ molecules

0.013 mole ×6.022×10²³ molecules/ 1 mol

0.078×10²³ molecules

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%