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 dа ta:

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