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24 May, 18:57

If 5.738 grams of AgNO3 is mixed with 4.115 grams of BaCl2 and allowed to react according to the balanced equation: BaCl2 (aq) + 2 AgNO3 (aq) → 2 AgCl (s) + Ba (NO3) 2 (aq) What is the limiting reagent? BaCl2AgNO3 How many grams of AgCl could be produced? grams AgCl What mass, in grams, of the excess reagent will remain? grams of excess reagent

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  1. 24 May, 20:15
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    Limiting reagent: AgNO3

    grams AgCl : 2.44 g AgCl

    grams of excess reagent remain: 0.62 g BaCl2

    Explanation:

    1. Change grams to mol:

    AgNO3:

    5.738g x (1mol/169.87g) = 0.034 mol AgNO3

    BaCl2:

    4.115g x (1 mol/208.23g) = 0.020 mol BaCl2

    2. Limiting reagent:

    AgNO3:

    0.034 mol AgNO3 x (1 mol BaCL2 / 2mol AgNO3) = 0.017 mol BaCl2

    BaCl2:

    0.020 mol BaCl2 x (2 mol AgNO3/1 mol BaCl2) = 0.04 mol AgNO3

    Limiting reagent: AgNO3

    3. Grams of AgCl produced:

    Using the limiting reagent:

    0.017 mol AgNO3 x (2mol AgCl / 2 mol AgNO3) = 0.017 mol AgCl

    4. Change mol to grams:

    0.017 mol AgCl x (143.32 g AgCl / 1mol AgCl) = 2.44 g AgCl

    5. Grams of the excess reagent:

    0.034 mol AgNO3 x (1 mol BaCl2 / 2 mol AgNO3) = 0.017 mol BaCl2

    0.020 mol BaCl2 - 0.017 mol BaCl2 = 0.003 mol BaCl2

    0.003 mol BaCl2 x (208.23 g BaCl2 / 1 mol BaCl2) = 0.62 g BaCl2
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