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 — Determine Mole Ratio from Balanced Equation

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

Step 2 — Calculate Moles of NaCl Produced

With a 1:1 stoichiometric ratio, 2.0 moles of Na will yield 2.0 moles of NaCl, assuming Cl₂ is in excess and Na is the limiting reactant.

Step 3 — Compute Molar Mass of NaCl

Using approximate standard atomic masses: Na = 22.99 g·mol⁻¹, Cl = 35.45 g·mol⁻¹.
The molar mass of NaCl is 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 — Apply Significant Figures for Reporting

• The input value 2.0 mol has 2 significant figures. • Molar masses (22.99 and 35.45) are typically precise to four significant figures; however, the final result should reflect the precision of the least precise measurement, which is 2.0 mol (2 s.f.). • Consequently, the theoretical yield can be reported as:

• Exact Calculation Result: 116.88 g
• Rounded to 3 s.f. (common practice): 117 g
• Rounded to 2 s.f. (consistent with input): 1.2 × 10² g (or 120 g)

Verification

• Mole ratio 1:1 confirms moles Na = moles NaCl (2.0 mol). • Calculation 2.0 × 58.44 ≈ 116.9 g is numerically consistent and dimensionally correct.