emanuel20971 emanuel20971

27-06-2022
Chemistry

contestada

How many moles FeBr3 are required
to generate 275 g NaBr?

2FeBr3 + 3Na₂S → Fe₂S3 +6NaBr

Respuesta :

nkosimlungisi908 nkosimlungisi908

27-06-2022

Answer:

0.893mol

Explanation:

n = m ÷ M

= 275 ÷ (23 + 80)

= 2.67 mol

* now use the Mol ratio *

NaBr : FeBr3

6 : 2

2.67 : x

5.67 = 6x

n( FeBr3 ) = 0.893 mol

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