The mass of HgS formed is approximately 2.31 grams.
To find the mass of HgS formed, we need to determine the limiting reactant in the reaction between Hg(NO3)2 and Na2S. The balanced chemical equation for the reaction is:
\[ \text{Hg(NO}_3)_2 (aq) + \text{Na}_2\text{S} (aq) \rightarrow \text{HgS} (s) + 2\text{NaNO}_3 (aq) \]
First, let's find the moles of each reactant:
1. Moles of Hg(NO3)2:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \]
\[ \text{Moles} = \frac{2.0 \, \text{g}}{200.59 \, \text{g/mol}} \approx 0.00998 \, \text{mol} \]
2. Moles of Na2S:
\[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \]
\[ \text{Moles} = \frac{2.0 \, \text{g}}{78.04 \, \text{g/mol}} \approx 0.0256 \, \text{mol} \]
Now, we need to find the limiting reactant. The balanced equation indicates a 1:1 molar ratio between Hg(NO3)2 and HgS. Therefore, the limiting reactant is the one that produces fewer moles of HgS.
In this case, Hg(NO3)2 produces fewer moles of HgS, so it is the limiting reactant.
Now, we can calculate the moles of HgS formed using the limiting reactant:
\[ \text{Moles of HgS} = \text{Moles of limiting reactant} \]
\[ \text{Moles of HgS} = 0.00998 \, \text{mol} \]
Finally, we can find the mass of HgS formed:
\[ \text{Mass} = \text{Moles} \times \text{Molar mass} \]
\[ \text{Mass} = 0.00998 \, \text{mol} \times 232.66 \, \text{g/mol} \approx 2.31 \, \text{g} \]
Therefore, the mass of HgS formed is approximately 2.31 grams.