Question

A sample of 211 g of iron (III) bromide is reacted with

Answer 1

FeBr₃ ⇒ limiting reactant

mol NaBr = 1.428

Further explanation

Reaction

2FeBr₃ + 3Na₂S → Fe₂S₃ + 6NaBr

Limiting reactant⇒ smaller ratio (mol divide by coefficient reaction)

• FeBr₃

211 g of Iron (III) bromide(MW=295,56 g/mol), so mol FeBr₃ :

$\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714$

• Na₂S

186 g of Sodium sulfide(MW=78,0452 g/mol), so mol Na₂S :

$\tt n=\dfrac{186}{78.0452}=2.38$

Coefficient ratio from the equation FeBr₃ :  Na₂S = 2 : 3, so mol ratio :

$\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793$

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

$\tt \dfrac{6}{3}\times 0.714=1.428$