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Aqueous silver nitrate reacts with aqeous potassium iodide in a double-replacement reaction to produce a precipitate of silver iodide. If 34.6 ml of 0.563 M silver nitrate are used with 148.4 ml of potassium iodide:

a) What molarity of potassium iodide will be needed to consume all of the silver nitrate?

b) What will be the molarity of the aqueous product?

Please show your work so I can do similar problems later on

  • chemistry -

    moles AgNO3 = M x L = 0.563M x 0.0346L = 0.020 moles approximately but you need to redo all of these since I estimate all of the numbers.

    KI(aq) + AgNO3(aq) ==>AgI(s) + KNO3(aq)
    Now convert moles AgNO3 to moles KI using the coefficients in the balanced equation.
    moles AgNO3 x (1 mole KI/1 mole AgNO3) = 0.02 x (1/1) = 0.02 mol KI.

    MKI = moles KI/L KI
    You know moles KI and L KI, solve for M KI.

    b) The aqueous product is KNO3. Use EITHER KI or AgNO3 (you don't need both) to convert to moles KNO3. The same procedure is used with the coefficients.
    moles AgNO3 x (1 mole KNO3/1 mole AgNO3) = 0.02 x (1 mole KNO3/1 mole AgNO3) = 0.02 x (1/1/) = 0.02 moles KNO3.
    Then M KNO3 = moles KNO3/L soln.
    YOu know moles KNO3 and you know the total volume of the solution.

  • chemistry -

    Thank you so much, I really appreciate your help!

  • chemistry -

    Sorry, but how did you find the total volume of the solution?

  • chemistry -

    34.6 mL AgNO3 + 148.4 mL KI = ? mL total.
    Don't get intimidated because these are mL or some unit you don't usually see. You add a gallon to a gallon and you have 2 gallons. Or a quart to a pint and you have 3 pints.

  • chemistry -

    Oh! Okay thank you, that's very good advice. I understand now

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