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

Md Ariful Islam

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

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

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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