Cl2 + 2Na → 2NaCl How many grams of NaCl are produced by the reaction of 0.300 L of chlorine gas at STP with excess sodium?

1.52 g of NaCl

Explanation:

We'll begin by calculating the number of mole of chlorine gas (Cl₂) that occupied 0.3 L at STP. This can be obtained as follow:

Recall: 1 mole of any gas occupy 22.4 L at STP.

1 mole of Cl₂ occupy 22.4 L at STP.

Therefore, Xmol of Cl₂ will occupy 0.3 L at STP i.e

Xmol of Cl₂ = 0.3 / 22.4

Xmol of Cl₂ = 0.013 mole

Thus, 0.013 mole of Cl₂ occupied 0.3 L at STP.

Next, we shall determine the number of mole of NaCl produced from the reaction.

Cl₂ + 2Na —> 2NaCl

From the balanced equation above,

1 mole of Cl₂ reacted to produce 2 moles of NaCl.

Therefore, 0.013 mole of Cl₂ will react to produce = 0.013 × 2 = 0.026 mole of NaCl.

Thus, 0.026 mole of NaCl is produced.

Finally, we shall determine the mass of the NaCl. This can be obtained as follow:

Mole of NaCl = 0.026 mol

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl =?

Mole = mass /Molar mass

0.026 = mass of NaCl / 58.5

Cross multiply

Mass of NaCl = 0.026 × 58.5

Mass of NaCl = 1.52 g

Therefore, 1.52 g of NaCl were produced from the reaction

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