Question

What is the theoretical yield of sodium chloride when 8.3 g of sodium is reacted with 4.5 g of chlorine? 2Na(s) + Cl2(g) → 2NaCl(s)

Answer 1

The theoretical yield of sodium chloride when 8.3 g of sodium react with 4.5 g of chlorine is 3.30 g

We'll begin calculating the masses of Na and Cl that reacted and the mass of NaCl produced from the balanced equation.

2Na + Cl₂ —> 2NaCl

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.

From the balanced equation above,

46 g of Na reacted with 71 g of Cl₂

Therefore,

8.3 g of Na will react with = (8.3 × 71)/46 = 12.81 g of Cl₂

From the calculation made above, we can see that a higher amount (i.e 12.81 g) of Cl₂ than what was given (i.e 4.5 g) is needed to react completely with 8.3 g of Na.

Therefore, Cl₂ is the limiting reactant and Na is the excess reactant

• Finally, we shall determine the theoretical yield of NaCl by using the limiting reactant (i.e Cl₂)

From the balanced equation above,

71 g of Cl₂ reacted to produce 117 g of NaCl.

Therefore,

4.5 g of Cl₂ will react to produce = (4.5 × 117)/71 = 3.30 g of NaCl

Thus, the theoretical yield of NaCl produced is 3.30 g

Learn more: brainly.com/question/25669079