Question

Calculate the enthalpy change for producing 215 g AlCl3(s) from its elements in their natural state using the following thermochemical equations.    Given: ∆Hºrxn

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2 -1049

HCl(g) → HCl(aq)    -73.5

½H2(g) + ½Cl2 → HCl(g) -92.5 (∆Hºf)

AlCl3(s) → AlCl3(aq) -323

Answer 1

First, claculate moles of AlCl3(s) required

mol = mass/MW = 215/133.34 = 1.612 mol of AlCl3 required

then...

divide  

2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2 -1049 by 2, since we need only 1 mol of Al(s)

Al(s) + 3HCl(aq) → AlCl3(aq) + 3/2H2 -1049/2 = -524.5 kJ

Now... add AlCl3(aq) since we need to cancel AlCl3(aq) , also add AlCl3(s) as product

Al(s) + 3HCl(aq) → AlCl3(aq) + 3/2H2 = - 524.5 kJ

AlCl3(aq) --> AlCl3(s)    = -(-323) = 323 kJ

add both

Al(s) + 3HCl(aq) → AlCl3(s) + 3/2H2   = - 524.5 +323 = -201.5 kJ

now... we need to change HCl to H2(g)

multiply by 3: and add

3HCl(g) → 3HCl(aq)    -73.5*3

Al(s) + 3HCl(aq) → AlCl3(s) + 3/2H2 -201.5 kJ

add

Al(s) + 3HCl(g ) → AlCl3(s) + 3/2H2 -201.5-220.5 = -422

Al(s) + 3HCl(g ) → AlCl3(s) + 3/2H2 -422 kJ

now... get rid o H2 via:

½H2(g) + ½Cl2 → HCl(g) -92.5 (∆Hºf)

multiply by 3

3/2H2(g) + 3/2Cl2 → 3HCl(g) -92.5*3= -277.5

now, add to:

Al(s) + 3HCl(g ) → AlCl3(s) + 3/2H2 -422

3/H2(g) + 3/2Cl2 → 3HCl(g) -277.5

add now

Al(s) + 3/2Cl2 (g ) → AlCl3(s) H = -422 -277.5 = -699.5 kJ