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
Now I go from grams to moles of each element?
2 answers
yes.
I have figured out that 32.34 grams of solid precipitate will form but I'm not sure how to figure out the grams of the reactant in excess? I thought it was 22.01 g but that isn't right.
1 answer