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. \( 58.5 \, \text{g} \)
• B. \( 116.9 \, \text{g} \)
• 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

The balanced chemical equation tells us that 2 moles of Na react with 1 mole of Cl\(_2\) to form 2 moles of NaCl. Therefore, the mole ratio between Na and NaCl is 1:1. Step 1: Find the molar mass of NaCl: - The molar mass of Na is \( 22.99 \, \text{g/mol} \), - The molar mass of Cl is \( 35.45 \, \text{g/mol} \). Thus, the molar mass of NaCl is: \[ 22.99 + 35.45 = 58.44 \, \text{g/mol} \] Step 2: Use the number of moles of sodium to find the mass of NaCl produced: Since the mole ratio is 1:1, 2.0 moles of Na will produce 2.0 moles of NaCl. Now, calculate the mass of NaCl: \[ \text{Mass of NaCl} = \text{moles of NaCl} \times \text{molar mass of NaCl} \] \[ \text{Mass of NaCl} = 2.0 \times 58.44 = 116.88 \, \text{g} \] Thus, the mass of sodium chloride produced is \( 58.5 \, \text{g} \).