Chemistry, 01.07.2019 19:20 kaylatunell123

If a solution containing 80.701 g80.701 g of mercury(ii) perchlorate is allowed to react completely with a solution containing 12.026 g12.026 g of sodium sulfide, how many grams of solid precipitate will form? precipitate: gg how many grams of the reactant in excess will remain after the reaction? excess reactant:

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The excess reagent for the given chemical reaction is mercury (II) perchlorate and the amount left after the completion of reaction is 0.048 moles. The amount of mercury sulfide formed in the reaction is 35.82 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For sodium sulfide:

Given mass of sodium sulfide = 12.026 g

Molar mass of sodium sulfide = 78.045 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium sulfide}=\frac{12.026g}{78.045g/mol}=0.154mol$

For mercury (II) perchlorate:

Given mass of mercury (II) perchlorate = 80.701 g

Molar mass of mercury (II) perchlorate = 399.49 g/mol

Putting values in above equation, we get:

$\text{Moles of mercury (II) perchlorate}=\frac{80.701g}{399.49g/mol}=0.202mol$

For the given chemical equation:

$Hg(ClO_4)_2(aq.)+Na_2S(aq.)\rightarrow HgS(s)+2NaClO_4(aq.)$

Here, the solid precipitate is mercury sulfide.

By Stoichiometry of the reaction:

1 mole of sodium sulfide reacts with 1 mole of mercury (II) perchlorate.

So, 0.154 moles of sodium sulfide will react with = $\frac{1}{1}\times 0.154=0.154moles$ of mercury (II) perchlorate

As, given amount of mercury (II) perchlorate is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium sulfide is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (mercury (II) perchlorate) left = 0.202 - 0.154 = 0.048 moles

By Stoichiometry of the reaction:

1 mole of sodium sulfide reacts with 1 mole of mercury sulfide.

So, 0.154 moles of sodium sulfide will react with = $\frac{1}{1}\times 0.154=0.154moles$ of mercury sulfide.

Now, calculating the mass of mercury sulfide from equation 1, we get:

Molar mass of mercury sulfide = 232.66 g/mol

Moles of mercury sulfide = 0.154 moles

Putting values in equation 1, we get:

$0.154mol=\frac{\text{Mass of mercury sulfide}}{232.66g/mol}\\\\\text{Mass of mercury sulfide}=35.82g$

Hence, the excess reagent for the given chemical reaction is mercury (II) perchlorate and the amount left after the completion of reaction is 0.048 moles. The amount of mercury sulfide formed in the reaction is 35.82 grams.

Answer from: Quest

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What city is located at 17° south 36° east answers: a)surfer’ town b)white’s beach c)red bank d)

Answer from: Quest

it's filtration, but this is an assumption so my apologies if i got this

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If a solution containing 80.701 g80.701 g of mercury(ii) perchlorate is allowed to react completely...

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