If a solution containing 33.82 g of mercury(II) acetate is allowed to react completely with a solution containing 9.718 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?
Write and balance the equation. Then work as far as you can (show your work) and I'll be glad to help you through.
C4H6O4Hg + Na2SO4 ----------->
2Na(C4H6O4)+ HgSO4
Can't you start? You almost ALWAYS must have mols to start.
33.82g/318.678= 0.106 mols Mercury ii acetate
9.718g/142.04= 0.0684 mols sodium sulfate
Since you have four significant figures in 33.82 AND in 9.718, I would have four places in the answer; i.e., add another place to each of the mols.
Next, using the coefficients in the balanced equation, convert mols EACH to mols HgSO4. The SMALLER number is the correct one and the reagent that produced that number is the limiting reagent.
Take that number mols and convert to grams.
i think my equation is wrong. should it be
C4H6O4Hg + Na2SO4 ----------->
Na(C4H6O4)+ HgSO4
but then there are no coefficients
acetic acid is CH3COOH and we often see it written as HC2H3O2.Therefore, mercuty(II) acetate is Hg(C2H3O2)2. You used the right molar mass for it of 318.
Hg(C2H3O2)2 + Na2SO4 ==> HgSO4 + 2NaC2H3O2
Of course you have coefficients. If there is nothing in front of a compound, the one (1) is always understood to be there. Remember x means 1x and 2x means two of that number.