What mass of iron is formed when 21.35 liters of aluminum react with iron (II) sulfate?

Jul 15, 2018

The question needs reproposal and a bit of work...

Explanation:

${\rho}_{\text{aluminum}} = 2.71 \cdot g \cdot c {m}^{-} 3$...here a VOLUME of $21.35 \cdot L$ is proposed...

And since $\rho = \text{mass"/"volume}$, $\text{mass} = 21.35 \cdot L \times 2.71 \cdot g \cdot c {m}^{-} 3 \times 1000 \cdot c {m}^{3} \cdot {L}^{-} 1 = 57858.5 \cdot g$...

But this aluminum is proposed to reduce ferrous sulfate...for which the following equation applies...

$2 A l \left(s\right) + 3 F e S {O}_{4} \left(s\right) \stackrel{\Delta}{\rightarrow} A {l}_{2} {\left(S {O}_{4}\right)}_{3} + 3 F e \left(s\right)$

And so ${n}_{\text{Al}} = \frac{57858.5 \cdot g}{26.98 \cdot g \cdot m o {l}^{-} 1} = 2144.5 \cdot m o l$...and given the stoichiometry of the equation, $\frac{3}{2} \cdot \text{equiv}$ of ferrous sulfate can be reduced....to give a mass of iron....

$\frac{3}{2} \times 2144.5 \cdot m o l \times 55.85 \cdot g \cdot m o {l}^{-} 1 = 180 \cdot \text{tonne}$...a rather large scale reaction....