Question

2 Cl2 + 4 NaOH = 3 NaCl + 1 NaClO2 + 2 H2O;11.9 g Cl2 is reacted with 12.0 g NaOH. Determine which is the limiting reactant.

Answer 1

Answer

NaOH is the limiting reactant

Explanation

Given reaction:

2 Cl₂ + 4 NaOH → 3 NaCl + 1 NaClO₂ + 2 H₂O

2 mol of Cl₂ = 71.0 g

4 mol of NaOH = 160.0 g

What to find:

The limiting reactant.

Step-by-step solution:

From the given reaction; 71.0 g of Cl₂ reacted with 160.0 g of NaOH

So the given reacting mass (11.9 g) Cl₂ is expected to react with:

$\frac{11.9\text{ g }Cl_2\times160.0\text{ g NaOH}}{71\text{ g }Cl_2}=26.8\text{ g NaOH}$

Note that the given mass of NaOH is 12.0 g but the actual mass needed to consume 11.9 g Cl₂ is 26.8 g. Hence, Cl₂ is the excess reactant and NaOH is the limiting reactant