# Q. An alloy of aluminium and copper was treated with aqueous `HCl.` The aluminium dissolved according to the reaction:   `Al + 3H^+ -> Al^(3+) +(3/2)H_2` but the copper remained as pure...

Q. An alloy of aluminium and copper was treated with aqueous `HCl.` The aluminium dissolved according to the reaction:

`Al + 3H^+ -> Al^(3+) +(3/2)H_2`

but the copper remained as pure metal. A 0.350-g sample of the alloy gave 415 cc of `H_2` measured at 273 K and 1 atm pressure.What is the weight percentage of `Al` in the alloy?

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

`Al+3H^+ rarr Al^(3+)+3/2H_2`

We can use the ideal gas equation to get the amount of `H_2` released.

`PV = nRT`

`P = 1atm`

`V = 415cc = 0.415L`

`R = 0.08206(atmL)/(molK)`

`T = 273K`

`n = (PV)/(RT)`

`n = (1xx0.415)/(0.08206xx273)`

`n = 0.0185mol`

Mole ratio

`Al:H_2 = 1:3/2`

Amount of Al moles reacted `= 0.0185/3xx2 = 0.012mol`

Molar mass of Al `= 27g/(mol)`

Mass of Al in alloy `= 0.012xx27 = 0.333g`

weight % of Al in alloy `= 0.333/0.35xx100% = 95.27%`

So the weight % of Al in alloy is 95.27%

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llltkl | College Teacher | (Level 3) Valedictorian

Posted on

Let the weight percentage of Aluminium in the given alloy be x.

So, 0.350 g sample of the alloy contains `(0.350*x)/100` , i.e. `0.0035x` g Al.

By the stoichiometry of the reaction of Al (the only metal reacting here),

26.98 g Al, on reaction with sufficient HCl, produces `3/2*22400` cc of `H_2` (g) at NTP (273K and 1 atm.)

Therefore, `0.0035x` g Al should produce `(3*22400*0.0035x)/(2*26.98)` cc of `H_2` (g)

By the condition of the problem,

`(3*22400*0.0035x)/(2*26.98)=415`

`rArr x=(415*2*26.98)/(3*22400*0.0035)=95.2`

Thus, the weight percentage of Al in the alloy was 95.2.

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