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Chemistry Question H

Chemistry
Tutor: None Selected Time limit: 0 Hours

If a solution containing 33.272 g of mercury(II) nitrate is allowed to react completely with a solution containing 10.872 g of sodium sulfate, how many grams of solid precipitate will be formed?

How many grams of the reactant will remain in excess 

Dec 2nd, 2014

Hg(NO3)2 + Na2SO4 --> 2NaNO3 + HgSO4

Hg(NO3)2: 200.59 + (14.01 + 16.00*3)*2 = 324.61 g/mol
Na2SO4: 22.99*2 + 14.01 + 16.00*3 = 107.99 g/mol
HgSO4: 200.59 + 32.07 + 16.00*4 = 296.66 g/mol

We have:
33.272 g Hg(NO3)2 * 1 mol Hg(NO3)2 / 324.61 g Hg(NO3)2 = .102 mol Hg(NO3)2
10.872 g Na2SO4 * 1 mol Na2SO4 / 107.99 g Na2SO4 = .101 mol Na2SO4

Ratio of Hg(NO3)2 to Na2SO4 is 1:1 from the balanced chemical equation, therefore Na2SO4 is the limiting reactant.

.101 mol Na2SO4 * 1 mol HgSO4 / 1 mol Na2SO4 * 296.66 g HgSO4 / 1 mol HgSO4 = 29.96 g HgSO4

Dec 2nd, 2014

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