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30 November, 11:49

How many grams of sodium chromate, Na2CrO4, are needed to react completely with

56.7g of silver nitrate, AgNO3?

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Answers (1)
  1. 30 November, 14:45
    0
    We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

    Explanation:

    Step 1: data given

    Mass of silver nitrate AgNO3 = 56.7 grams

    Molar mass AgNO3 = 169.87 g/mol

    Molar mass of Na2CrO4 = 161.97 g/mol

    Step 2: The balanced equation

    2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

    Step 3: Calculate moles AgNO3

    Moles AgNO3 = mass AgNO3 / molar mass AgNO3

    Moles AgNO3 = 56.7 grams / 169.87 g/mol

    Moles AgNO3 = 0.334 moles

    Step 4: Calculate moles Na2CrO4 moles needed

    For 2 moles AgNO3 we need 1 mol Na2CrO4 to produce 1 mol Ag2CrO4 and 2 moles NaNO3

    For 0.334 moles AgNO3 we need 0.334 / 2 = 0.167 moles Na2CrO4

    Step 5: Calculate mass Na2CrO4

    Mass Na2CrO4 = 0.167 moles * 161.97 g/mol

    Mass Na2CrO4 = 27.0 grams

    We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3
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