contestada

A sample of 211 g of iron (III) bromide is reacted with
186 g of sodium sulfide to produce iron (III) sulfide
and sodium bromide. Using the balanced equation
below, predict which is the limiting reactant and the
maximum amount in moles of sodium bromide that
can be produced.
2FeBr3 + 3Na2S → Fe2S3 + 6NaBr

Respuesta :

ardni313

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 :

[tex]\tt n=\dfrac{mass}{MW}\\\\n=\dfrac{211}{295,56}\\\\n=0.714[/tex]

  • Na₂S

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

[tex]\tt n=\dfrac{186}{78.0452}=2.38[/tex]

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

[tex]\tt FeBr_3\div Na_2S=\dfrac{0.714}{2}\div \dfrac{2.38}{3}=0.357\div 0.793[/tex]

So  FeBr₃ as a limiting reactant(smaller ratio)

mol NaBr based on limiting reactant (FeBr₃) :

[tex]\tt \dfrac{6}{3}\times 0.714=1.428[/tex]