Can someone please help me with the following problem?
1.Calculate the volume of 1.88 M NaI that would be needed to precipitate all of the Hg+2 ion from 482 mL of a 1.61 M Hg(NO3)2. The equation for the reaction is
thank you!
Hg^+2 + 2I^- ==> HgI2
How much Hg^+2 do you have? That's M x L = 1.61 M x 0.482 L = ??
Looking at the equation, it will take twice as many moles of iodide.
Then M x L = moles Iodide.
You know M and you know moles, calculate liters.