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) :

Answer:
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Answer:
15.89grams
Explanation:
The mass of a substance can be calculated from it's mole value by using the formula:
mole = mass/molar mass
According to this question, there are 0.250 moles of copper. Hence, the mass of copper can be calculated as follows:
Molar mass of Cu = 63.55g/mol
0.250 = mass/63.55
mass = 0.250 × 63.55
mass = 15.8875
Mass of Cu in 0.250mol is 15.89grams.
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