Respuesta :
Answer:
There will remain 0.7 moles of O2 in excess
Explanation:
Step 1: The balanced equation
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
Step 2: Given data
Number of moles of ZnS = 3.0 mol
Number of moles of O2 = 5.2 mol
Molar mass ZnS = 97.47 g/mol
Molar mass of O2 =32 g/mol
Step 3: Calculate limiting reactant
For 2 moles of ZnS consumed, we need 3 moles of O2 to produce 2 moles of ZnO and 2 moles of SO2
ZnS is the limiting reactant. It will completely be consumed. ( 3.0 moles). O2 is the reactant in excess. There will be consumed 3.0 mol * 3/2 = 4.5 mol
There will remain 5.2 - 4.5 = 0.7 moles of O2
There will be produced 3 moles of ZnO and 3 moles of SO2
The amount (in moles) of the excess reactant remaining from the reaction is 0.7 mole
Balanced equation
2ZnS + 3O₂ → 2ZnO + 2SO₂
From the balanced equation above,
2 moles of ZnS reacted with 3 moles of O₂
How to determine the excess reactant
From the balanced equation above,
2 moles of ZnS reacted with 3 moles of O₂
Therefore,
3 moles of ZnS will react with = (3 × 3) / 2 = 4.5 moles of O₂
Thus, only 4.5 moles out of 5.2 moles of O₂ is needed to reacted completely with 3 moles of ZnS.
Therefore, ZnS is the limiting reactant and O₂ is the excess reactant.
How to determine the amount remaining
- Mole of O₂ given = 5.2 moles
- Mole of O₂ that reacted = 4.5 moles
- Amount of O₂ remaining =?
Amount of O₂ remaining = 5.2 – 4.5
Amount of O₂ remaining = 0.7 mole
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