posted by Chelsea on .
If a solution containing 52.044 g of mercury(II) chlorate is allowed to react completely with a solution containing 15.488 g of sodium sulfate, how many grams of solid precipitate will be formed? How many grams of the reactant in excess will remain after the reaction?
I have the balanced equation right I think...
Hg(ClO3)2+Na2SO4--->HgSO4(s) + 2NaClO3
I have figured out that there are 32.34 g of solid precipitate but I don't know how to figure out the grams of the reactant in excess remaining.?
LR = limiting reagaent
ER = excess reagent.
I didn't check any of your numbers but if they are right here is what you do.
Use the coefficients to tell you just as you determined the mols of HgSO4 formed.
mols ER used = mols LR used x coefficients to convert LR to ER.
Then mols ER x molalr mass ER = g ER used.
mols ER initially - mols ER used = mols ER remaining unreacted.