Write the reaction between solutions of mercury (II) nitrate and sodium sulfide.
Hg(NO3)2 + Na2S ==> HgS + 2NaNO3
a) how many moles of sodium nitrate form from the reaction of 2.85 mol of sodium sulfide?
2.85 mols Na2S x (2 mols NaNO3/1 mol Na2S) = ? mols NaNO3
b) how many litres of 0.150 mol/L mercury (II) nitrate react with 0.540 L of 0.653 mol/L sodium sulfide?
mols Na2S = M x L = 0.653 M x 0.540 L = ?
mols Hg(NO3)2 = mols Na2S. Same reasoning as part a.
Then M Hg(NO3)2 = mols/L. You know mols and M, solve for L
c) how many grams of precipitate would you get from the reaction of 1.00 L of 0.550 mol/L sodium sulfide with mercury (II) nitrate?
mols Na2S = M x L = 0.550 x 1.00 L = 0.550 mols.
You get 1 mol HgS(s) for every mol Na2S so mols HgS = mols Na2S using the same reasoning as part a.
Then grams = mols HgS x molar mass = ? You know mols HgS and molar mass HgS so substitute and solve for grams.
Post work if you get stuck. Be specific if you don't understand one of the steps above.
I’ve tried everything but just can’t get it.
Write the reaction between solutions of mercury (II) nitrate and sodium sulfide.
a) how many moles of sodium nitrate form from the reaction of 2.85 mol of sodium sulfide?
b) how many litres of 0.150 mol/L mercury (II) nitrate react with 0.540 L of 0.653 mol/L sodium sulfide?
c) how many grams of precipitate would you get from the reaction of 1.00 L of 0.550 mol/L sodium sulfide with mercury (II) nitrate?
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