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19 March, 23:42

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

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  1. 20 March, 03:35
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    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
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