chem

25.0 mL of 0.125 M silver nitrate are mixed with 35.0mL of 0.100 M sodium sulfate to give solid silver sulfate. Write the net ionic equation, and determine the mass in grams of silver sulfate produced.

I think the net ionic equation is
2Ag(aq) + SO4^-2(aq) ---> AgSO4(s)

but not sure what to do next...

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asked by Lori
  1. M=mol/L

    "25.0 mL of 0.125 M silver nitrate" = (25/1000 L)(0.125 mol/L)

    "35.0mL of 0.100 M sodium sulfate"=
    (35/1000 L)(0.100 mol/L)

    now that you have the moles of Ag and SO4^2- convert to moles of AgSO4 through the mole ratio and smaller value is the limiting reagent

    take the limiting reagent and multiply by the molar mass AgSO4

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  2. I think silver sulfate is Ag2SO4.
    Also I think this problem fosters the idea that Ag2SO4 is relatively insoluble and it isn't. In fact, about 0.250 g of Ag2SO4 will dissolve in this solution and that amount will be lost.

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  3. you're right doc its Ag2SO4, thanks guys!

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    posted by Wade

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