Calculate the mass of Na2S needed if a solution containing 2g of Hg(NO3)2 was added to Na2S solution. ( Hg= 200.59, N= 14, O= 16, Na= 23, S=32)​

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) =

2b2t

explanation:

2b2t

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