Answer to Question #269664 in Chemistry for Grace

Question

Identify the limiting reactant when 1.7 g of sodium reacts with 2.6 l of chlorine gas at stp to produce sodium chloride?

Answer 1

Solution:

Calculate the moles of each reactant:

The molar mass of sodium (Na) is 23 g/mol

Hence,

(1.7 g Na) × (1 mol Na / 23 g Na) = 0.074 mol Na

At STP, one mole of any gas occupies a volume of 22.4 L.

Hence,

(2.6 L Cl₂) × (1 mol Cl₂ / 22.4 L Cl₂) = 0.116 mol Cl₂

Balanced chemical equation:

2Na(s) + Cl₂(g) → 2NaCl(s)

According to stoichiometry:

2 mol of Na reacts with 1 mol of Cl₂

Thus, 0.074 mol of Na reacts with:

(0.074 mol Na) × (1 mol Cl₂ / 2 mol Na) = 0.037 mol Cl₂

However, initially there is 0.116 mol of Cl₂ (according to the task).

Therefore, Na acts as limiting reactant and Cl₂ is excess reactant.

Answer: Sodium (Na) acts as limiting reactant.