Suppose that 2.00mL of 0.02 mol/L aqueous sodium sulphide, Na2S is used to test a 60.0mL sample of water containing 0.0004 mol/L mecury (II) nitrate Hg(NO3)2, ions.
a. Write the balanced equation
Hg(NO3)2 + Na2S = HgS + 2NaNO3
b. Determine the limiting reactant
Na2S = 2.00mL/1000mL x 0.02mol/L
= 0.00004 mol
Hg(NO3)2 = 60.0mL/1000mL x 0.0004mol
= 0.000024 mol
limiting reactant is the mercury nitrate.
c. calculate the maximum mass of precipitate formed.
what is the precipitate?
1 answer
You need to learn the solubility rules. HgS is the ppt (the solid).
1 answer