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FeBr₃ ⇒ limiting reactant
mol NaBr = 1.428
<h3>Further explanation</h3>
Reaction
2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr
Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)
211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :
186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :
Coefficient ratio from the equation FeBr₃ : Na₂S = 2 : 3, so mol ratio :
So FeBr₃ as a limiting reactant(smaller ratio)
mol NaBr based on limiting reactant (FeBr₃) :
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<h3>
Answer:</h3>
Balanced equation: 4Fe + 3O₂ → 2Fe₂O₃
Moles of oxygen gas = 9 moles
<h3>
Explanation:</h3>
To answer the question;
- We first write the balanced equation between iron metal and Oxygen
- The balanced equation is given as;
4Fe + 3O₂ → 2Fe₂O₃
- We are given 6 moles of Fe₂O₃
We are required to determine the number of moles of oxygen needed to form 6 moles of Fe₂O₃.
- From the equation, 3 moles of oxygen gas reacts to produce 2 moles of Fe₂O₃
- This means, the mole ratio of O₂ to Fe₂O₃ is 3 : 2
Therefore; Moles of O₂ = Moles of Fe₂O₃ × 3/2
Hence, moles of oxygen = 6 moles × 3/2
= 9 moles
Thus, Moles of Oxygen needed is 9 moles
