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 $2FeBr_3 + 3Na_2S -> Fe_2S_3 + 6NaBr$ ?

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Answer 1 · 2017-08-15T07:34:00

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

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....

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

How many moles of "iron(Ill) sulfide""iron(Ill) sulfide", Fe_2S_3Fe_2S_3, would be produced from the complete reaction of 449*g449*g "iron(III) bromide""iron(III) bromide" in the reaction 2FeBr_3 + 3Na_2S -> Fe_2S_3 + 6NaBr2FeBr_3 + 3Na_2S -> Fe_2S_3 + 6NaBr?