If a solution containing 118.08 g of mercury(II) chlorate is allowed to react completely with a solution containing 16.642 g of sodium sulfide, how many grams of solid precipitate will form

49.544 g.

Explanation:

The balanced equation of reaction is given below;

Hg(ClO3)2 (aq) + Na2S (aq) > 2 NaClO3 (aq) + HgS (s).

So, the parameters given in the question are; Mass of Hg(ClO3) = 118.08 and the mass of Na2S = 16.642 g.

Therefore, the first thing we are going to be looking at is the reactant which is the limiting reagent from the number of moles

(1). For Hg(ClO3), the molar mass = 367.5 g/mol. Therefore, the number of moles, n = mass/ molar mass.

Number of moles = 118.08/ 367.5.

Number of moles = 0.321 moles.

(2). For Na2S, the molar mass = 78.05 g /mol.

The number of moles = 16.643 / 78.05.

The number of moles= 0.213.

Therefore, the limiting reagent = Na2S.

This means that the excess reagent is Hg(ClO3).

==> In the balanced equation of reaction above, the solid precipitate = HgS.

Hence, the mass of HgS formed = 0.213 × 232.6 g/mol. = 49.544 g.

the electronic configuration of an atom with eight electrons is 1s²2s²2p⁴. the orbital diagrams are shown in the figure.

as per the hunds rule:

1. electrons tend to avoid being in the same orbital as per as practicable.

2. electrons in different orbitals of the same energy will have their spin parallel.

this can be illustrated from the multiplicity of the orbitals of the atom. as there are four electrons in the p orbital and p orbital has 3 sub-shells. the electrons can be fed in different ways as shown in the figure but the correct orbital diagram will be as per the hunds rule.

just did the assignment and the answer was #1 +2 n