Answer:
1. Cl₂ is the limiting
2. 8.24 g of NaCl.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2Na + Cl₂ —> 2NaCl
Next, we shall determine the masses of Na and Cl₂ that reacted and the mass of NaCl produced from balanced equation. This is illustrated below:
Molar mass of Na = 23 g/mol
Mass of Na from the balanced equation = 2 × 23 = 46 g
Molar mass of Cl₂ = 2 × 35.5
= 71 g/mol
Mass of Cl₂ from the balanced equation = 1 × 71 = 71 g
Molar mass of NaCl = 23 + 35.5
= 58.5 g/mol
Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g
SUMMARY:
From the balanced equation above,
46 g of Na reacted with 71 g of Cl₂ to produce 117 g of NaCl.
1. Determination of the limiting reactant.
From the balanced equation above,
46 g of Na reacted with 71 g of Cl₂.
Therefore, 12 g of Na will react with = (12 × 71) / 46 = 18.52 g of Cl₂.
From the calculation made above, we can see clearly that a higher amount (i.e 18.52 g) of Cl₂ than what was given (i.e 5 g) is needed to react completely with 12 g of Na.
Therefore, Cl₂ is the limiting reactant and Na is the excess reactant.
2. Determination of the amount of NaCl produced from the reaction.
In this case the limiting reactant will be used because it will give the maximum yield of NaCl as all of it consumed in the reaction.
The limiting reactant is Cl₂ and the mass of NaCl produced can be obtained as follow:
From the balanced equation above,
71 g of Cl₂ reacted to produce 117 g of NaCl.
Therefore, 5 g of Cl₂ will react to produce = (5 × 117)/71 = 8.24 g of NaCl.
Thus, 8.24 g of NaCl were obtained from the reaction.
