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

1 Answer
Jul 15, 2018

Answer:

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....