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23 November, 09:59

A 3.90 g sample of Cl2 reacts with PCl3 to form 11.45 g of PCl5 according to the reaction below. How much PCl3 is needed? [Hint: You do not need to calculate moles to answer.]

PCl3 + Cl2 - > PCl5 Amount of PCl3:

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  1. 23 November, 10:10
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    We need 7.55 grams of PCl3

    Explanation:

    Step 1: Data given

    Mass of Cl2 = 3.90 grams

    Mass of PCl5 = 11.45 grams

    Molar mass Cl2 = 70.9 g/mol

    Molar mass PCl5 = 208.24 g/mol

    Step 2: The balanced equation

    PCl3 + Cl2 → PCl5

    Step 3: calculate moles

    Moles = mass / molar mass

    Moles Cl2 = 3.90 grams / 70.9 g/mol

    Moles Cl2 = 0.0550 moles

    Moles PCl5 = 11.45 grams / 208.24 g/mol

    Moles PCl5 = 0.0550 moles

    Step 4: Calculate moles PCl3

    For 1 mol PCl3 we need 1 mol Cl2 to produce 1 mol PCl5

    For 0.0550 moles PCl5 we will need 0.0550 moles Cl2 and 0.0550 moles PCl3

    Step 5: Calculate mass PCl3

    Mass PCl3 = moles * molar mass

    Mass PCl3 = 0.0550 moles * 137.33 g/mol

    Mass PCl3 = 7.55 grams

    Without calculating the number of moles:

    Mass of Cl2 + mass PCl3 = mass PCl5

    Mass PCl3 = 11.45 - 3.90 = 7.55 grams

    We need 7.55 grams of PCl3
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