An alloy of aluminium and copper was treated with aqueous HCl. The aluminium dissolved according to the reaction:
Al + 3H+ -------> Al3+ + 3/2 H,
but the copper remained as pure metal. A 0.350-g samole of the alloy gave 415cc of H2 measured at 273K and 1 atm pressure. What is the weight percentage of Al in the alloy?
Let us, simplify the reaction by multiplying with two, as
[Al + 3H+ ---> Al3+ + 3/2 H2 ] x 2
or, 2Al + 6H+ ---> 2Al3+ + 3H2
The molar mass of Al = 27
Here, we see that, 2 moles of Al gives 3 moles of H2 gas(molar mass=2)
Since, the reaction is taken at STP (273K and 1 atm pressure)
So, 0.350 gm of sample releases 415ml (=415cc) of hydrogen gas.
So, at STP , a 1gm of sample will release = 415/0.350 = 1185.7 ml of hydrogen gas.
Now from question, at STP, 1mole of a gas = 22.4 L
So, 3 moles of hydrogen gas = 22.4 x3 = 67.2L =67200ml
So, we can say that, 67200 ml of Hydrogen gas is released by 54 gm of Al in the alloy.
So, 1185.7 ml of hydrogen gas will require = 54/67200 x1185.7 = 0.95 gm of Al of the alloy.
Hence, we see that 0.350 gm sample contains 0.95 gm Al
So, each gram of sample will contain = 0.95/0.350= 2.71 gm of Al
So, the weight percentage of Al in every 100 gm of sample = weight of Al/ total weight x100 %= 2.71/100 x 100= 2.71 %