Question

One gram of an alloy of aluminium and magnesium when heated with excess of dil. Hcl forms magnesium chloride, aluminium chloride and hydrogen, the evolved hydrogen collected over mercury at of 0degree Celsius has a volume of 1.12 liters at 1 atm pressure. Calculate the composition of the allow.

Answer 1

The ionic reaction equation for the reaction is :

1.) Mg(s) + 2H⁺ (aq) —> Mg²⁺(aq) + H₂(g)

Mole ratio is 1 : 2

2.) 2Al(s) + 6H⁺(aq) —> 2Al³⁺(aq) + 3H₂(g)

Mole ratio is 1 : 3

1 mole of H₂ = 22.4 litres at stp

1.12 / 22.4 = 0.05 moles.

Moles of Mg = 0.05 / 2 = 0.025 moles

Mass is :

0.025 × 24 = 0.6g

Moles of Al = 0.05 / 3 = 0.01667 moles.

Mass is :

0.01667 × 27 = 0.45 g

Composition of the alloy :

0.45 + 0.6 = 1.05g

Answer 2

Answer:

Explanation:

let the mole of magnesium be x and mole of aluminium be y

Mg + 2Hcl---MgCL2 +H2

Al +3Hcl ----AlCl3 +1.5H2

NOW

24x +27y=39

x +1.5y=2

solving these two equation we get

x=.5 ,y=1

percentage of magnesium=30.76%

percentage of lauminium=69.24%