Question

50. 3 g of an alloy of aluminium and magnesium was
treated with dilute HCl (excess). Aluminium and
magnesium react with HCl as per following
reaction :
2AI + 6HCI 2AlCl3 + 3H2
Mg + 2HCl → MgCl2 + H2
The hydrogen produced has a volume of 3.136 L
at STP. The percentage composition of
aluminium by mass in the alloy is (Mg = 24 g/mol,
Al = 27 g/mol)​

Answer 1

Answer:

hey, it should be 39 g of alloy!

Explanation:

let the mole of magnesium be x and mole of aluminum 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 aluminum=69.24%

applying PV=nRT for hydrogen

n=2 mole

let the mole of magnesium be x and mole of aluminum 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 aluminium=69.24%