Question

A 10.0 g piece of pure aluminum is placed in 75.0 mL of 0.54 M hydrochloric acid at STP condition. They react as follows: 2Al + 6HCl → 3H2(g) + 2AlCl3 Calculate the following:

a. Volume, in liters, of hydrogen gas.

b. Molarity of Al+3. (Assume 75.0 mL solution.)

c. Molarity of Cl–. (Assume 75.0 mL solution.)

Answer 1

Answer – We are given, mass of Al = 10.0 g

Molarity of HCl = 0.54 M, volum e of HCl = 75.0 mL

Reaction – 2Al + 6HCl ------> 3H₂(g) + 2AlCl_­3

Now first we need to calculate the moles of each reactant –

Moles of Al = 10.0 g / 26.982 g.mol^-1

                     = 0.371 moles

Moles of HCl = 0.54 M * 0.075 L

                       = 0.0405 moles

Now we need to find out limiting reactant

Moles of H₂ from the Al

From the balanced equation –

2 moles of Al = 3 moles of H₂

So, 0.371 moles of Al = ?

= 0.556 moles H₂

Moles of H₂ from the HCl

From the balanced equation –

6 moles of HCl = 3 moles of H₂

So, 0.0405 moles of HCl = ?

= 0.0203 moles H₂

So the moles of H₂ got lowest from the HCl, so limiting reactant is HCl.

moles of H₂ = 0.0203 moles

a)Volume of the H₂ gas

so, at STP, P = 1.0 atm, T = 273 K,

we already calculated moles of H₂ , so using the ideal gas law

PV= nRT

V = nRT/P

     = 0.0203 moles * 0.0821 L.atm.mol^-1.K^-1*273 K / 1.0 atm

      = 0.454 L

b) Now we need to calculate moles of AlCl₃ from the HCl

From the balanced equation -

From the balanced equation –

6 moles of HCl = 2 moles of AlCl₃

So, 0.0405 moles of HCl = ?

= 0.0135 moles AlCl₃

We know,

1 moles of AlCl₃ = 1 moles of Al³⁺

So, 0.0135 moles of AlCl₃ = ?

= 0.0135 moles of Al³⁺

[Al³⁺] = 0.0135 moles / 0.075 L

           = 0.180 M

1 moles of Al³⁺ = 3 moles of Cl^-

So, moles of Cl^- = 3*0.0135 moles

                            = 0.0405 moles

[Cl^-] = 0.0405 mole / 0.075 L

          = 0.540 M