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

1 Answer
anor277
Nov 25, 2017

You gots the stoichiometric reaction.....

Explanation:

#2FeBr_3 + 2Na_2S rarr Fe_2S_3 + 6NaBr#

....the which tells you that treatment of approx. #600*g# #"ferric bromide"# upon treatment with approx. #160*g# #"sodium sulfide"# to give #208*g# #"ferric sulfide"#, and an equivalent mass of #"sodium bromide...."#

We have a molar quantity of #(449*g)/(207.9*g*mol^-1)=2.16*mol# with respect to #"ferric bromide"#, and given the stoichiometry, we get a #1.08*mol# quantity of #"ferric sulfide"#, i.e. a mass of ...

#1.08*molxx207.9*g*mol^-1=??*g#

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