Answer:
1.433g of HgS are produced
Explanation:
A Solution Containing 2.0 Grams Of Hg(NO3)2 Was Added To A Solution Containing 2.0 Grams Of Na2S. Calculate the mass of the HgS that was formed (it is a precipitate) according to this reaction:
Based on the reaction:
Na2S + Hg(NO3)2 → HgS + 2NaNO3
To solve this question we need to find the moles of each reactant in order to find the limiting reactant. The moles of limiting reactant = moles of HgS:
Moles Na2S -Molar mass: 78.0452 g/mol-
2.0g * (1mol / 78.0452g) = 0.0256 moles Na2S
Moles Hg(NO3)2 -324.7g/mol-
2.0g * (1mol / 324.7g) = 0.006159 moles Hg(NO3)2
As the reaction is 1:1, and moles of Hg(NO3)2 < moles Na2S
The moles of Hg(NO3)2 = Moles HgS = 0.006159 moles
The mass is:
Mass HgS -Molar mass: 232.66g/mol-:
0.006159 moles * (232.66g/mol) =
