Question

Dissolve 7, 8 grams of Mg and Al mixture with excess HCl solution. After the reaction, the mass of acid solution increased by 7, 0 grams. Determine the mass of each metal in the mix.​

Answer 1

Mg = 2,5 g; Al = 5,3 g

Step-by-step explanation:

1) Reactions

Mg + 2HCl ⟶ MgCl₂ + H₂

2Al + 6HCl ⟶ 2AlCl₃ + 3H₂

2) Mass of each metal

If there had been no reaction, the mass of the solution would have increased by 7,8 g.

The mass increased by only 7,0 g.

The missing 0,8 g must represent the mass of the hydrogen generated by the reaction.

We have two relations:

Mass of Mg + mass of Al = 7,8 g

H₂ from Mg + H₂ from Al = 0,8 g

i) Calculate the moles of H₂

`n = \text{0,8 g H}_(2) * \frac{\text{1 mol H}_(2)}{\text{2,016 g H}_(2)} = \text{0.40 mol H}_(2)`

(ii) Solve the relationship

Let x = mass of Mg. Then

7,8 - x = mass of Al

Moles of Mg = x/24.30

Moles of Al = (7,8 - x)/26.98

Moles of H₂ from Mg = (1/1) × moles of Mg = 1 × (x/24,30) = 0,0412x

Moles of H₂ from Al = (3/2) × Moles of Al = 1.5(7,8 - x)/26,98 = (11,7 -1,5x)/26,98

`\begin{array}{rcl}0,0412x + ( 11,7 - 1,5x )/(26,98) &= &0,40\\\\ 1,11x + 11.7 - 1,5x &=& 10.7\\-0.39x& = &-1.0\\x &=& \text{2.5 g}\\\end{array}`

Mass of Mg = 2,5 g

Mass of Al = 7,8 g - 2,5 g = 5,3 g

The masses of the metals are Mg = 2,5 g; Al = 5,3 g