2 Na + Cl2 --> 2 NaCl According to the balanced equation above, how many Lof Cl2 at STP are
needed to completely react with 25.0 g Na?

12.208 L

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.