How many liters of hydrogen can be produced at a pressure of 2 atm and a temperature of 298 K

1.17 L of H₂

Explanation:

We'll begin by calculating the number of mole in 2.3 g of Mg. This can be obtained as follow:

Mass of Mg = 2.3 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /molar mass

Mole of Mg = 2.3 / 24

Mole of Mg = 0.096 mole

Next, we shall determine the number of mole of H₂ produced by the reaction of 2.3 g (i.e 0.096 mole) of Mg. This can be obtained as follow:

Mg + 2HCl —> MgCl₂ + H₂

From the balanced equation above,

1 mole of Mg reacted to 1 mole of H₂.

Therefore, 0.096 mole of Mg will also react to produce 0.096 mole of H₂.

Finally, we shall determine volume of H₂ produced from the reaction. This can be obtained as follow:

Number of mole (n) of H₂ = 0.096 mole

Pressure (P) = 2 atm

Temperature (T) = 298 K

Gas constant (R) = 0.0821 atm.L/Kmol

Volume (V) of H₂ =?

PV = nRT

2 × V = 0.096 × 0.0821 × 298

Divide both side by 2

V = (0.096 × 0.0821 × 298) /2

V = 1.17 L

Therefore, 1.17 L of H₂ were obtained from the reaction.