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 just learnt this recently but i need to look through my notes again

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i