Answer:
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.
