Question
250 mL of 2 x 10-5 mol/L AgNO3 is mixed with 250 mL 2 x 10-4 mol/L NaI. What is the mass of the precipitate?
Answers
DrBob222
AgNO3 + NaI ==> AgI + NaNO3
millimoles AgNO3 = mL x M = 500E-5
millimoles NaI = 500E-4
So you will form 500E-5 millimoles AgI
Convert that to moles AgI and multiply by molar mass AgI.
Post your work if you get stuck.
millimoles AgNO3 = mL x M = 500E-5
millimoles NaI = 500E-4
So you will form 500E-5 millimoles AgI
Convert that to moles AgI and multiply by molar mass AgI.
Post your work if you get stuck.