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29 August, 18:37

How many grams of iron (II) chloride are needed to produce 44.3 g iron (II) phosphate in the presence of excess sodium phosphate?

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  1. 29 August, 18:42
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    47.2 g

    Explanation:

    Let's consider the following double displacement reaction.

    3 FeCl₂ + 2 Na₃PO₄ → Fe₃ (PO₄) ₂ + 6 NaCl

    The molar mass of Fe₃ (PO₄) ₂ is 357.48 g/mol. The moles corresponding to 44.3 g are:

    44.3 g * (1 mol / 357.48 g) = 0.124 mol

    The molar ratio of Fe₃ (PO₄) ₂ to FeCl₂ is 1:3. The moles of FeCl₂ are:

    3 * 0.124 mol = 0.372 mol

    The molar mass of FeCl₂ is 126.75 g/mol. The mass of FeCl₂ is:

    0.372 mol * (126.75 g/mol) = 47.2 g
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