Question

Mercury can be obtained by reacting mercury(ii) sulfide with calcium oxide. how many grams of calcium oxide are needed to produce 31.28 g of hg?

Answer 1

`8.75~g~CaO`

Step-by-step explanation:

First we have to start with the reaction:

`HgS~+~CaO~->~Hg~+~CaS~+~CaSO_4`

The next step is to balance the reaction, we can start with Oxygen, so:

`HgS~+~4CaO~->~Hg~+~CaS~+~CaSO_4`

Then we can continue with Ca:

`HgS~+~4CaO~->~Hg~+~3CaS~+~CaSO_4`

Then we can balance S:

`4HgS~+~4CaO~->~Hg~+~3CaS~+~CaSO_4`

And finally with Hg:

`4HgS~+~4CaO~->~4Hg~+~3CaS~+~CaSO_4`

With the balance reaction we know that the molar ratio between Hg nd CaO is 4:4. Therefore, the nex step is the conversion of 31.28 g Hg to moles of Hg using the atomic mass of Hg (200.59 g/mol).

`31.28~g~Hg(1~mol~Hg)/(200.59~g~Hg)`

`0.156~mol~Hg`

The next step, using the molar ratio (4:4) and the molar mass of CaO (56.1 g/mol) we can calculate the grams of CaO that we need:

`0.156~mol~Hg(4~mol~CaO)/(4~mol~Hg)(56.1~g~CaO)/(1~mol~CaO)`

`8.75~g~CaO`

Answer 2

Reaction between mercury (ii) sulfide with Calcium oxide is as shown;
4 HgS +4 CaO = 4 Hg + 3 CaS + CaSO4
The molar mass of HgS is 232.66 g/mol
The number of moles = 31.28/232.66
= 0.1344 moles
The mole ratio of HgS : CaO is 1: 1
Therefore, the number of moles of Calcium oxide is 0.1344 moles
1 mole of CaO Contains 56 g/mol
Thus, the mass of CaO is 56 g/mol
= 0.1344 moles × 56 g/mol
= 7.5264 g