February 24, 2017

Homework Help: Chemistry

Posted by Chelsea on Sunday, October 13, 2013 at 7:45pm.

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.?

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