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

b

explanation:

mitosis

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-