Answer:
7.30g Oâ‚‚ are in excess and 21.3g of SOâ‚‚ should be produced.
Explanation:
The reaction of Zinc sulfide (ZnS) with oxygen (Oâ‚‚) produce ZnO and SOâ‚‚ as follows:
2ZnS + 3O₂ → 2ZnO + 2SO₂
Using this reaction we can find the amount of SOâ‚‚ produced converting the ZnS and Oâ‚‚ to moles and finding limiting and excess reactant as follows:
Moles ZnS -Molar mass: 97.47g/mol-:
32.5g * (1mol / 97.47g) = 0.333 moles ZnS
Moles Oâ‚‚ -Molar mass: 32g/mol-:
23.3g Oâ‚‚ * (1mol / 32g) = 0.728 moles Oâ‚‚
For a complete reaction of the 0.333 moles of ZnS are required:
0.333 moles ZnS * (3mol Oâ‚‚ / 2mol ZnS) = 0.500 moles Oâ‚‚.
As there are 0.728 moles of Oâ‚‚, limiting reactant is ZnS and excess reactant is Oâ‚‚.
The moles and mass of Oâ‚‚ in excess are:
0.728mol - 0.500mol = 0.228moles Oâ‚‚ * (32g / mol) = 7.30g Oâ‚‚ are in excess
The produced mass of SOâ‚‚ -Obtained from the moles of ZnS- is:
0.333 moles of ZnS * (2mol SOâ‚‚ / 2mol ZnS) = 0.333 moles SOâ‚‚
0.333 moles SOâ‚‚ * (64.066g / mol) =
21.3g of SOâ‚‚ should be produced
