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30 May, 00:46

If 25 grams of NaI is mixed with excess amount Pb (NO3) 2, what would be the product?

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  1. 30 May, 04:46
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    If 25 grams of NaI is mixed with an excess amount of Pb (NO3) 2 we get 38.45 grams of PbI2 and 14.18 grams of NaNO3

    Explanation:

    Step 1: Data given

    Mass of NaI = 25.00 grams

    Pb (NO3) 2 is in excess

    Step 2: The balanced equation

    2NaI + Pb (NO3) 2 → PbI2 + 2NaNO3

    Step 3: Calculate moles of NaI

    Moles NaI = mass NaI / molar mass NaI

    Moles NaI = 25.00 / 149.89 g/mol

    Moles NaI = 0.1668 moles

    Step 4: Calculate moles of PbI2 and NaNO3

    For 2 moles of NaI we need 1 mol of Pb (NO3) 2 to produce 1 mol of PbI2 and 2 moles of NaNO3

    For 0.1668 moles of of NaI we will have 0.1668/2 = 0.0834 moles of PbI2 and 0.1668 moles of NaNO3

    Step 5: Calculate mass of the products

    Mass PbI2 = 0.0834 * 461.01 g/mol

    Mass PbI2 = 38.45 grams

    Mass NaNO3 = 0.1668 mol * 84.99 g/mol

    Mass NaNO3 = 14.18 grams

    If 25 grams of NaI is mixed with an excess amount of Pb (NO3) 2 we get 38.45 grams of PbI2 and 14.18 grams of NaNO3
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