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?

The mass of mercury (II) sulfide is 53.51 g

Explanation:

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

     .....(1)

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:

Given mass of sodium sulfide = 17.796 g

Molar mass of sodium sulfide = 78.04 g/mol

Putting values in equation 1, we get:

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

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 = 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 = 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:

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

i'm pretty sure its b i might be wrong though!