Question

Question was provided via image upload.

Explanation:

Step1: Balance the reaction

First, balance the chemical equation:

$$2\text{Na} + \text{Cl}_2 ightarrow 2\text{NaCl}$$

Step2: Find moles of Na

Calculate moles of Na from mass:
$$n_{\text{Na}} = \frac{45\ \text{g}}{22.99\ \text{g/mol}} = 1.957\ \text{mol}$$

Step3: Moles of NaCl from Na

Use mole ratio (2:2) to find NaCl moles:
$$n_{\text{NaCl (from Na)}} = 1.957\ \text{mol} \times \frac{2}{2} = 1.957\ \text{mol}$$

Step4: Mass of NaCl from Na

Convert moles to grams (molar mass of NaCl = $22.99 + 35.45 = 58.44\ \text{g/mol}$):
$$m_{\text{NaCl (from Na)}} = 1.957\ \text{mol} \times 58.44\ \text{g/mol} = 114.4\ \text{g}$$

Step5: Find moles of Cl₂

Calculate moles of Cl₂ from mass:
$$n_{\text{Cl}_2} = \frac{35\ \text{g}}{70.90\ \text{g/mol}} = 0.4937\ \text{mol}$$

Step6: Moles of NaCl from Cl₂

Use mole ratio (1:2) to find NaCl moles:
$$n_{\text{NaCl (from Cl}_2\text{)}} = 0.4937\ \text{mol} \times \frac{2}{1} = 0.9874\ \text{mol}$$

Step7: Mass of NaCl from Cl₂

Convert moles to grams:
$$m_{\text{NaCl (from Cl}_2\text{)}} = 0.9874\ \text{mol} \times 58.44\ \text{g/mol} = 57.7\ \text{g}$$

Step8: Identify limiting reactant

The reactant that produces less product ($\text{Cl}_2$) is limiting. The theoretical yield is the smaller mass of NaCl.

Answer:

Balanced reaction: $2\text{Na} + \text{Cl}_2
ightarrow 2\text{NaCl}$
Theoretical yield of NaCl: $57.7\ \text{g}$
Limiting reactant: $\text{Cl}_2$