Question

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

Answer 1

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

Explanation:

`rho_"aluminum"=2.71*g*cm^-3`...here a VOLUME of `21.35*L` is proposed...

And since `rho="mass"/"volume"`, `"mass"=21.35*Lxx2.71*g*cm^-3xx1000*cm^3*L^-1=57858.5*g`...

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

`2Al(s) + 3FeSO_4(s)stackrel(Delta)rarr Al_2(SO_4)_3 + 3Fe(s)`

And so `n_"Al"=(57858.5*g)/(26.98*g*mol^-1)=2144.5*mol`...and given the stoichiometry of the equation, `3/2*"equiv"` of ferrous sulfate can be reduced....to give a mass of iron....

`3/2xx2144.5*molxx55.85*g*mol^-1=180*"tonne"`...a rather large scale reaction....