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
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