Question

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

2FeBr3 + 3Na₂S → Fe₂S3 +6NaBr

Answer 1

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