reaction enthalpy for the formation of aluminum chloride?

Question

can someone please help me and explain how to get the answer? the first correct answer gets a best answer. thanks a lot!

Calculate the reaction enthalpy for the formation

2 Al(s) + 3 Cl2(g) − → 2 AlCl3(s) ,

of anhydrous aluminum chloride using the data

2 Al(s) + 6 HCl(aq) − → 2 AlCl3(aq) + 3 H2(g) ∆H◦ = −1049 kJ

HCl(g) −→ HCl(aq) ∆H◦ = −74.8 kJ

H2 (g) + Cl2 (g) −→ 2 HCl(g) ∆H◦ = −185 kJ

AlCl3(s) −→ AlCl3(aq) ∆H◦ = −323 kJ

1. −1450.85 kJ

2. −1826.2 kJ

3. −1502.4 kJ

4. −1883.5 kJ

5. −1225.7 kJ

6. −1100.36 kJ

7. −1406.8 kJ

Answer 1

Favorite Answer

2 Al(s) + 6 HCl(aq) >> 2 AlCl3(aq) + 3 H2(g)…|……-1049

+ 6 x HCl(g) >> HCl(aq)…………………………|…+ 6x (-74.8)

+ 3 x H2(g) + Cl2(g) >> 2 HCl(g).………………|…+ 3x (-185)

- 2 x AlCl3(s) >> AlCl3(aq)………………………|…- 2x (-323)

————————————————————————————

2 Al(s) + 6 HCl(aq) + 6 HCl(g) + 3 H2(g) + 3 Cl2(g) - 2 AlCl3(s)

…>>…

2 AlCl3(aq) + 3 H2(g) + 6 HCl(aq) + 6 HCl(g) - 2 AlCl3(aq)

…

reducing (?) delivers:

2 Al(s) + 3 Cl2(g) >> 2 AlCl3(s)

.……………………= -1049 - 448.8 - 555 + 646 = -1406.8 kJ

Source(s): Just what I learned during studies

Answer 2

Anhydrous Aluminum Chloride

Source(s): https://shrinke.im/a7Wms

Answer 3

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1) 2 Al (s) + 3/2 O2 (g) ----> Al2O3 (s) 2) C2H5OH (l) + 3 O2 (g) ------> 2 CO2(g) + 3 H20(l) 3) NaOH (aq) + HCl (aq) -------> NaCl (aq) + H2O (l) 4) 2 C (s) + 3/2 H2 (g) + 1/2 Cl2 (g) -------> C2H3Cl (g) 5) C6H6 (l) + 7 1/2 O2 (g) ---> 6 CO2 (g) + 3 H2O (l) 6) NH4Br (s) + aq -------> NH4+ (aq) + Br- (aq)