Answer:
Cl₂ is the limiting reactant.
Explanation:
Firstly, we need to write the balanced reaction:2Na + Cl₂ → 2NaCl,
It is clear that 2 mol of Na react with 1 mol of Cl₂ to produce 2 mol of NaCl.
Firstly, we need to calculate the no. of moles of 8.0 g Na and 8.0 g Cl₂:For Na:
n = mass/molar mass = (8.0 g)/(22.989 g/mol) = 0.348 mol.
For Cl₂:
n = mass/molar mass = (8.0 g)/(70.9 g/mol) = 0.113 mol.
From the stichiometry: 2 mol of Na react with 1 mol of Cl₂ with (2: 1) molar ratio.The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it.
So, 0.113 mol of Cl₂ react completely with 0.226 mol of Na which be in excess (0.384 - 0.226 = 0.158 mol).
So, the limiting reactant is Cl₂ which is the reactant has the lowest molar ratio.
Diatomic chlorine (Cl₂) is the limiting reactant for the reaction between 8 g of sodium and 8 g of diatomic chlorine.
The limiting reactant in a chemical reaction is simply defined as the reactant which is completely used up in the chemical reaction.
With the above information in mind, we can determine the limiting reactant for the reaction between 8 g of sodium and 8 g of diatomic chlorine (Cl₂). This can be obtained as follow:
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 = 35.5 × 2 = 71 g/mol
Mass of Cl₂ from the balanced equation = 1 × 71 = 71 g
From the balanced equation above,
46 g of Na reacted with 71 g of Cl₂
Therefore,
8 g of Na will react with = [tex]\frac{8 * 71}{46}\\\\[/tex] = 12.3 g of Cl₂
From the calculation made above, we can see clearly that a higher mass of Cl₂ (i.e 12.3 g) than what was given (i.e 8 g) is needed to react completely with 8 g of Na.
Therefore, Cl₂ is the limiting reactant and Na is the excess reactant.
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