Question

If a solution containing 53.99 g 53.99 g of mercury(II) nitrate is allowed to react completely with a solution containing 9.718 g 9.718 g of sodium sulfide, how many grams of solid precipitate will form?

Answer 1

Answer:

28.97 g

Explanation:

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

For mercury(II) nitrate

Mass = 53.99 g

Molar mass = 324.7 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{53.99\ g}{324.7\ g/mol}$

$Moles= 0.16627\ mol$

For sodium sulfide

Mass = 9.718 g

Molar mass = 78.0452 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{9.718\ g}{78.0452\ g/mol}$

$Moles= 0.12451\ mol$

According to the given reaction:

$Hg(NO_3)_2+Na_2S\rightarrow HgS+2NaNO_3$  

1 mole of mercury(II) nitrate reacts with 1 mole of sodium sulfide

So,  

0.16627 mole of mercury(II) nitrate reacts with 0.16627 mole of sodium sulfide

Moles of sodium sulfide = 0.16627 moles

Available moles of sodium sulfide = 0.12451 moles

Limiting reagent is the one which is present in small amount. Thus, sodium sulfide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of sodium sulfide produces 1 mole of mercury sulfide (precipitate)

So,  

0.12451 mole of sodium sulfide produces 0.12451 mole of mercury sulfide (precipitate)

Moles of mercury sulfide  = 0.12451 mole

Molar mass of mercury sulfide  = 232.66 g/mol

Mass of mercury sulfide  = Moles × Molar mass = 0.12451 × 232.66 g = 28.97 g

28.97 g  of solid precipitate will form.