Question

How many grams of sodium chloride (NaCl) are produced when 2.0 moles of sodium (Na) react with excess chlorine gas (Cl\(_2\)? The balanced chemical equation for the reaction is: \[ 2 \, \text{Na} (s) + \text{Cl}_2 (g) \rightarrow 2 \, \text{NaCl} (s) \]

• A. \( 116.9 \, \text{g} \)
• B. \( 58.5 \, \text{g} \)
• C. \( 117.0 \, \text{g} \)
• D. \( 231.5 \, \text{g} \)

Hint:

In stoichiometry, always use the mole ratios from the balanced chemical equation to convert between substances. Don’t forget to calculate the molar mass of compounds to find the mass of the product.

Answer 1

The Correct Option is A

Step 1 — Identify the mole ratio from the balanced equation

From the balanced equation: 2 Na + Cl₂ → 2 NaCl. The mole ratio between Na and NaCl is 2 mol Na : 2 mol NaCl, which simplifies to 1 : 1.

Step 2 — Convert given moles of Na to moles of NaCl

Because the stoichiometric ratio is 1:1, 2.0 moles Na will produce 2.0 moles NaCl (Cl₂ is in excess so Na is the limiting reagent).

Step 3 — Calculate the molar mass of NaCl

Use standard atomic masses (approximate): Na = 22.99 g·mol⁻¹, Cl = 35.45 g·mol⁻¹. 
Molar mass of NaCl = 22.99 + 35.45 = 58.44 g·mol⁻¹.

Step 4 — Convert moles of NaCl to grams

Mass = (moles) × (molar mass) = 2.0 mol × 58.44 g·mol⁻¹ = 116.88 g.

Step 5 — Significant figures and final reporting

• The given quantity 2.0 mol has 2 significant figures. • Molar masses (22.99 and 35.45) are typically given to four significant figures, but the final result should generally reflect the least precise measured value (here 2.0 has 2 s.f.). • Therefore you may report the theoretical yield as:

• Exact (from calculation): 116.88 g
• Rounded to 3 s.f. (common): 117 g
• Rounded to 2 s.f. (consistent with 2.0 mol): 1.2 × 10² g (or 120 g)

Quick check

• Mole ratio 1:1 ⟹ moles Na = moles NaCl (2.0 mol). • 2.0 × 58.44 ≈ 116.9 g — consistent and dimensionally correct.