posted by Kelsey on .
2.00mL of 0.01 mol/L aqueous sodium sulphide is used to test a 50.omL sample of water containing 0.0005 mol/L mercury (II) nitrate ions. What mass of precipitate is formed?
Na2S + Hg(NO3)2 --- 2NaNo3 + HgS
determine limiting reactant
using limiting reactant, calculate the masimum mass of precipitate formed
mol Na2S = (0.01mol/L)*2.00ml*1L/1000mL = 0.002mol
mol Hg(NO3)2 = 50.0ml*(0.0005mol/L)*1L/1000mL = 0.000025=2.5*10^-5
Hg(NO3)2 limiting reagent
max mass of precipitate = 2.5*10^-5 * molar mass of Hg(NO3)2
I thought I was to use a formula n=cXv. I had NaS
V >01 mol/L
Therefore I got NaS as .00002. So I had Na S as the limiting reagent
moles = M x L
For Na2S, 0.01 x 0.002 = 2 x 10^-5 moles
moles Hg(NO3)2 =
0.0005 x (50/1000)= 2.5 x 10^-5 moles
Limiting reagent is Na2S. (The set up for Na2S by bun is correct but math is not; answer is 2 x 10^-5 moles and not 2 x 10^-3 moles.