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

Md Ariful Islam

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Md Ariful Islam

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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....

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

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