Question

1gram alloy of aluminium and magnesium reacted with hcl and formed agcl3 mgcl2 and evolved hydrogen gas . H2 was contained in a container with volume 1200ml at 0 centigrade . Find the composition of aluminium and magnesium in the alloy

Answer 1

According to the following equation, for an ideal gas,
PV = nRT

​The reaction is at STP, which means temperature = 273.15 K and pressure = 1atm,

​n = PV/RT = 1 x 1.2 / 0.082 x 298 = 0.0491 moles
The following reactions take place,
Mg + 2 HCl → MgCl₂ + H₂
​2Al + 6HCl → 2AlCl₃ + 3H₂
Let Al be x and Mg be (1 - x)
Number of moles of Al = Mass / Molar Mass = x/27
Number of moles of Mg = (1 - x)/24
From the above equations, it is clear that,
2 moles of Al produces 3 moles of H₂ which implies that x/27 moles of Al produces (x/27) x (3/2) moles of H₂ = x/18
Also 1 mole of Mg gives 1 mole of H₂ . Hence, (1-x)/24 moles of Mg gives (1-x)/24 moles of H₂.

So, total moles of H₂ = x/18 + (1-x)/24 which is equal to 0.0491 moles
On solving, we get
x = 0.55 g
Hence, Amount of Al = 0.55 g
Amount of Mg = 1 - 0.55 = 0.45 g