Question

When aluminum is placed in concentrated hydrochloric acid, hydrogen gas is produced. 2 Al ( s ) + 6 HCl ( aq ) ⟶ 2 AlCl 3 ( aq ) + 3 H 2 ( g ) What volume of H 2 ( g ) is produced when 2.80 g Al ( s ) reacts at STP?

Answer 1

3.49 Liters of hydrogen is produced when 2.80 grams of aluminum reacts at STP.

Step-by-step explanation:

`2Al ( s ) + 6HCl ( aq )\rightarrow 2AlCl_3 ( aq) +3H_2 ( g )`

Moles of aluminum =
`(2.80 g)/(27 g/mol)=0.1037 mol`

According to reaction , 2 moles of aluminium gives 3 moles of hydrogen gas.Then 0.1037 moles of aluminum will give:

`(3)/(2)* 0.1037 mol=0.1556 mol` of hydrogen gas

Moles of hydrogen gas = n = 0.1556 mol

Pressure of the gas, at STP , P = 1 atm

Temperature of the gas, at STP = T = 273 K

Volume of the gas , At STP = V

`PV=nRT` ( Ideal gas equation )

`V=(nRT)/(P)=(0.1556 mol* 0.08210 atm L/mol K* 273 K)/(1 atm)=3.49 L`

3.49 Liters of hydrogen is produced when 2.80 grams of aluminum reacts at STP.