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

the answer is b. dilution