Question

What is the limiting reactant in a reaction that produces sodium chloride from 8 g of sodium and 8 g of diatomic chlorine?

Answer 1

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.