Question

1. cl₂ + 2 nabr → 2 nacl + br₂ if 25.0 g of sodium chloride is produced, what mass of cl₂ reacted?

Explanation:

Step1: Balance the chemical equation

The unbalanced equation is \( \text{Cl}_2 + \text{NaBr}
ightarrow \text{NaCl} + \text{Br}_2 \).
To balance it, we need 2 moles of \( \text{NaBr} \) to produce 2 moles of \( \text{NaCl} \) and 1 mole of \( \text{Br}_2 \) from 1 mole of \( \text{Cl}_2 \). So the balanced equation is:
\( \text{Cl}_2 + 2\text{NaBr}
ightarrow 2\text{NaCl} + \text{Br}_2 \)

Step2: Calculate moles of \( \text{NaCl} \)

Molar mass of \( \text{NaCl} \) is \( 58.44 \, \text{g/mol} \) (sodium: \( 22.99 \, \text{g/mol} \), chlorine: \( 35.45 \, \text{g/mol} \), so \( 22.99 + 35.45 = 58.44 \, \text{g/mol} \)).
Moles of \( \text{NaCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{25.0 \, \text{g}}{58.44 \, \text{g/mol}} \approx 0.4278 \, \text{mol} \)

Step3: Determine moles of \( \text{Cl}_2 \) from stoichiometry

From the balanced equation, the mole ratio of \( \text{Cl}_2 \) to \( \text{NaCl} \) is \( 1:2 \).
So moles of \( \text{Cl}_2 = \frac{1}{2} \times \) moles of \( \text{NaCl} = \frac{1}{2} \times 0.4278 \, \text{mol} \approx 0.2139 \, \text{mol} \)

Step4: Calculate mass of \( \text{Cl}_2 \)

Molar mass of \( \text{Cl}_2 \) is \( 70.90 \, \text{g/mol} \) (since \( 35.45 \times 2 = 70.90 \, \text{g/mol} \)).
Mass of \( \text{Cl}_2 = \text{moles} \times \text{molar mass} = 0.2139 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 15.17 \, \text{g} \)

Answer:

The mass of \( \text{Cl}_2 \) reacted is approximately \( 15.2 \, \text{g} \) (or more precisely \( \approx 15.17 \, \text{g} \))