What is the theoretical yield of NaBr
when 2.36 moles of FeBr3 reacts?
2FeBr3 + 3Na2S → Fе₂S3 + 6NaBr
[?] mol NaBr
Round your answer to the hundredths place

What is the theoretical yield of NaBr when 236 moles of FeBr3 reacts 2FeBr3 3Na2S FеS3 6NaBr mol NaBr Round your answer to the hundredths place class=

Respuesta :

The theoretical yield of NaBr given that 2.36 moles of FeBr₃ reacts is 7.08 moles

Balanced equation

2FeBr₃ + 3Na₂S → Fе₂S₃ + 6NaBr

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

How to determine the theoretical yield of NaBr

From the balanced equation above,

2 moles FeBr₃ reacted to produce 6 moles of NaBr

Therefore,

2.36 moles FeBr₃ will react to produce = (2.36 × 6) / 2 = 7.08 moles of NaBr

Therefore,

Thus, the theoretical yield of NaBr is 7.08 moles

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