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.

Question

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

but not sure what to do next...

Answer 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

Answer 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.

Answer 3

you're right doc its Ag2SO4, thanks guys!