How many moles of #"iron(Ill) sulfide"#, #Fe_2S_3#, would be produced from the complete reaction of #449*g# #"iron(III) bromide"# in the reaction #2FeBr_3 + 3Na_2S -> Fe_2S_3 + 6NaBr#?

1 Answer
Aug 15, 2017

Answer:

The stoichiometric equation specifies half an equiv of ferric sulfide with respect to ferric bromide.

Explanation:

And thus we simply calculate the moles of ferric bromide.

#"Moles of ferric bromide"=(449*g)/(295.96*g*mol^-1)=1.517*mol#.

And thus we gets an half equiv of the sulfide salt....which represents a mass of....

#1.517*cancel(mol)xx1/2xx207.90*g*cancel(mol^-1)~=150*g#....