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?

Nov 25, 2017

You gots the stoichiometric reaction.....

Explanation:

$2 F e B {r}_{3} + 2 N {a}_{2} S \rightarrow F {e}_{2} {S}_{3} + 6 N a B r$

....the which tells you that treatment of approx. $600 \cdot g$ $\text{ferric bromide}$ upon treatment with approx. $160 \cdot g$ $\text{sodium sulfide}$ to give $208 \cdot g$ $\text{ferric sulfide}$, and an equivalent mass of $\text{sodium bromide....}$

We have a molar quantity of $\frac{449 \cdot g}{207.9 \cdot g \cdot m o {l}^{-} 1} = 2.16 \cdot m o l$ with respect to $\text{ferric bromide}$, and given the stoichiometry, we get a $1.08 \cdot m o l$ quantity of $\text{ferric sulfide}$, i.e. a mass of ...

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