Question

If a solution containing 126.27 g of mercury(ii) nitrate is allowed to react completely with a solution containing 17.796 g of sodium sulfide, how many grams of solid precipitate will be formed?

Answer 1

The number of solid precipitate that will be formed is  37.08 g

calculation

write the equation for reaction

=Hg(NO3)2 +Na2SO4 = HgSO4(s) +2NaNO3(aq)

find the moles of each reactant

moles ofHg(NO3)2=126.27/324.6= 0.389 moles

moles of Na2SO4=17.796/142=0.125 moles

NaSO4 is the limiting reagent and by use of mole ratio of NaSO4:HgSO4 which is 1:1 therefore the moles of H2SO4 is also= 0.125 moles

mass HgSO4=moles x molar mass

=0.125 x296.65= 37.08g

Answer 2

Answer: The mass of mercury (II) sulfide is 53.51 g

Explanation:

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

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

• For mercury (II) nitrate:

Given mass of mercury (II) nitrate = 126.27 g

Molar mass of mercury (II) nitrate = 324.7 g/mol

Putting values in equation 1, we get:

$\text{Moles of mercury (II) nitrate}=\frac{126.27g}{324.7g/mol}=0.388mol$

• For sodium sulfide:

Given mass of sodium sulfide = 17.796 g

Molar mass of sodium sulfide = 78.04 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium sulfide}=\frac{17.796g}{78.04g/mol}=0.23mol$

The chemical equation for the reaction of Mercury (II) nitrate and sodium sulfide follows:

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

By Stoichiometry of the reaction:

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

So, 0.23 moles of sodium sulfide will react with = $\frac{1}{1}\times 0.23=0.23mol$ of mercury (II) nitrate

As, given amount of mercury (II) nitrate 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.

By Stoichiometry of the reaction:

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

So, 0.23 moles of sodium sulfide will produce = $\frac{1}{1}\times 0.23=0.23moles$ of mercury (II) sulfide

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

Molar mass of mercury (II) sulfide = 232.66 g/mol

Moles of mercury (II) sulfide = 0.23 moles

Putting values in equation 1, we get:

$0.23mol=\frac{\text{Mass of mercury (II) sulfide}}{232.66g/mol}\\\\\text{Mass of mercury (II) sulfide}=53.51g$

Hence, the mass of mercury (II) sulfide is 53.51 g