Al (s) + HCl (aq) - -> AlCl3 (aq) + H2 (g)

a. Balance the eqaution

b. determine the oxidation state of each atom on each side of the equation

c. If 1.562 g of Al reacts completely, how many grams of AlCl3 will be formed?

d. Which species is reduced and which is oxidized?

e. If the reaction is carried out in 240.0 mL of acid solution, what is the concentration of AlCl3, in M, when the reaction is complete?

f. How many moles of H2 gas will evolve during this reaction?

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

b) oxidation state of Al = 0

oxidation state of H in HCl = + 1

oxidation state of Cl in HCl = - 1

oxidation state of Al IN AlCl₃ = + 3

oxidation state of Cl in AlCl₃ = - 1

oxidation state of H in H₂ = 0

c).562 gm of Al

= 1.562 / 27

=.0578 mole of Al will form. 0578 mole of AlCl₃

mol weight of AlCl₃ = 27 + 35.5 x 3 = 133.5 gm

.0578 mole of AlCl₃ =.0578 x 133.5

= 7.71 gm.

d) Al is oxidised as it oxidation no increases and H is reduced as its oxidation no is reduced.

e) volume =.24 liter

concentration of AlCl₃ = (.0578 /.24) M

=.24 M

f) moles of H₂ gas evolved

= 3 / 2 x. 0578

=.0867 moles.