- Calculate the number of moles of sodium: \(n(\text{Na}) = \frac{35.0 \text{ g}}{22.99 \text{ g/mol}} \approx 1.5224 \text{ mol}\).
- Use the stoichiometric ratio from the balanced equation (\(2 \text{Na} : 1 \text{Cl}_2\)) to find the moles of chlorine required: \(n(\text{Cl}_2) = \frac{1.5224 \text{ mol}}{2} \approx 0.7612 \text{ mol}\).
- Convert the moles of chlorine to mass using its molar mass (\(70.90 \text{ g/mol}\)): \(0.7612 \text{ mol} \times 70.90 \text{ g/mol} \approx 53.969 \text{ g}\).
- Round the result to three significant figures to obtain the final answer: \(\boxed{54.0 \text{ g}}\).
Explanation
- Analysis of the Reaction Stoichiometry
To determine the mass of chlorine gas (\(\text{Cl}_2\)) required to react completely with \(35.0 \text{ g}\) of sodium metal (\(\text{Na}\)), we begin by examining the balanced chemical equation for the reaction:
\[2 \text{Na} + \text{Cl}_2 \rightarrow 2 \text{NaCl}\]
This equation indicates that 2 moles of sodium react with 1 mole of chlorine gas to produce 2 moles of sodium chloride. Our task is to perform stoichiometric calculations to find the required mass of chlorine gas based on the initial mass of sodium provided.
- Converting Mass to Moles of Sodium
First, we convert the mass of sodium given into moles using its molar mass. The molar mass of sodium is approximately \(22.99 \text{ g/mol}\).
The number of moles of sodium is calculated as:
\[n(\text{Na}) = \frac{\text{mass}(\text{Na})}{\text{molar mass}(\text{Na})} = \frac{35.0 \text{ g}}{22.99 \text{ g/mol}} \approx 1.5224 \text{ mol}\]
- Determining Moles of Chlorine Gas Required
According to the balanced chemical equation, the stoichiometric ratio between \(\text{Na}\) and \(\text{Cl}_2\) is \(2:1\). This means that 1 mole of \(\text{Cl}_2\) is required for every 2 moles of \(\text{Na}\). Therefore, the number of moles of \(\text{Cl}_2\) required is:
\[n(\text{Cl}_2) = \frac{1}{2} \times n(\text{Na}) = \frac{1.5224 \text{ mol}}{2} \approx 0.7612 \text{ mol}\]
- Calculating the Final Mass of Chlorine Gas
Now, we convert the moles of \(\text{Cl}_2\) back to mass. The molar mass of \(\text{Cl}_2\) is calculated from the atomic mass of chlorine, which is approximately \(35.45 \text{ g/mol}\). Thus, the molar mass of \(\text{Cl}_2\) is:
\[\text{Molar mass}(\text{Cl}_2) = 2 \times 35.45 \text{ g/mol} = 70.90 \text{ g/mol}\]
The required mass of \(\text{Cl}_2\) is then:
\[\text{mass}(\text{Cl}_2) = n(\text{Cl}_2) \times \text{molar mass}(\text{Cl}_2) = 0.7612 \text{ mol} \times 70.90 \text{ g/mol} \approx 53.969 \text{ g}\]
Considering the input \(35.0 \text{ g}\) has three significant figures, we round the final answer to three significant figures, which is \(54.0 \text{ g}\).
Examples
In real-life chemistry, determining exact stoichiometric ratios is vital for industrial processes like salt production, water treatment, or even cooking. For example, when creating a pickling brine, knowing exactly how much salt is required for a specific volume of water and vegetable weight ensures the correct osmotic balance for preservation. By applying molar mass and stoichiometric calculations as done here, chemists and engineers ensure efficiency and minimize waste in large-scale reactions.