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

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

Answer 2

Answer:

$8.75~g~CaO$

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).

$31.28~g~Hg\frac{1~mol~Hg}{200.59~g~Hg}$

$0.156~mol~Hg$

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:

$0.156~mol~Hg\frac{4~mol~CaO}{4~mol~Hg}\frac{56.1~g~CaO}{1~mol~CaO}$

$8.75~g~CaO$