Answer:
1.52 g of NaCl
Step-by-step 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