Answer:
87.75%
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 the balanced equation. This can be obtained as follow:
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.
Next, we shall determine the limiting reactant.
This can be obtained as follow:
From the balanced equation above,
46 g of Na reacted with 71 g of Cl₂.
Therefore, 2.6 g of Na will react with = (2.6 × 71)/46 = 4.01 g of Cl₂.
From the calculations made above, we can see that only 4.01 g of Cl₂ at of 5 g given in question reacted completely with 2.6 g of Na. Therefore, Na is the limiting reactant and Cl₂ is the excess reactant.
Next, we shall determine the theoretical yield of NaCl. The limiting reactant will be used to obtain the theoretical yield since all of it is consumed in the reaction.
The limiting reactant is Na and the theoretical yield of NaCl can be obtained as follow:
From the balanced equation above,
46 g of Na reacted to produce 117 g of NaCl.
Therefore, 2.6 g of Na will react to produce = (2.6 × 117)/46 = 6.61 g of NaCl.
Thus the theoretical yield of NaCl is 6.61 g
Finally, we shall determine the percentage yield of NaCl. This can be obtained as follow:
Actual yield of NaCl = 5.8 g
Theoretical yield of NaCl = 6.61 g
Percentage yield =?
Percentage yield = Actual yield /Theoretical yield × 100
Percentage yield = 5.8/6.61 ×100
Percentage yield of NaCl = 87.75%
