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 3.60 g Al ( s ) reacts at STP?

Answer 1

Answer:

The volume of hydrogen gas produced at STP is 4.90 liters.

Explanation:

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

Moles of aluminium =$\frac{3.60 g}{27 g/mol}=0.1333 mol$

According to reaction , 2 moles of aluminium gives 3 moles of hydrogen gas.

Then 0.1333 moles of aluminium will give:

$\frac{3}{2}\times 0.1333 mol=0.2 moles$ of hydrogen gas

Volume of 0.2 moles of hydrogen gas at STP = V

Temperature at STP = T = 298.15 K

Pressure at STP = P = 1 atm

n = 0.2 mol

PV = nRT (Ideal gas equation)

$V=\frac{nRT}{P}=\frac{0.2mol\times 0.0821 atm L/mol K\times 298.15 K}{1 atm}=4.8956 L\approx 4.90 L$

The volume of hydrogen gas produced at STP is 4.90 liters.