Answer:
12.208 L
Step-by-step explanation:
We'll begin by calculating the number of mole in 25 g of Na. This can be obtained as follow:
Mass of Na = 25 g
Molar mass of Na = 23 g/mol
Mole of Na =?
Mole = mass /Molar mass
Mole of Na = 25/23
Mole of Na = 1.09 moles
Next, we shall determine the number of mole of Cl2 required to react with 1.09 moles of Na. This can be obtained as follow:
2Na + Cl2 –> 2NaCl
From the balanced equation above,
2 moles of Na reacted with 1 mole of Cl2.
Therefore, 1.09 moles of Na will react with = (1.09 × 1)/2 = 0.545 mole of Cl2.
Thus, 0.545 mole of Cl2 is needed for the reaction.
Finally, we shall determine the volume of Cl2 required for the react as follow:
Recall: 1 mole of any gas occupy 22.4 L at STP.
1 mole of Cl2 occupied 22.4 L at STP.
Therefore, 0.545 mole of Cl2 of Cl2 will occupy = 0.545 × 22.4 = 12.208 L at STP.
Therefore, 12.208 L of Cl2 is needed for the reaction.