- Identify the metals involved: Sodium (\(Na\)) and Potassium (\(K\)).
- Consult the activity series: Potassium (\(K\)) is more reactive than Sodium (\(Na\)) because it appears earlier in the series.
- Determine the outcome: A more reactive metal displaces a less reactive one, so Potassium will displace Sodium from its compound.
- Select the product: The resulting reaction is \(K_2CO_3 + Na\).
- Final Answer: \(\boxed{K_2CO_3 + Na}\)
Explanation
- Analyzing the Reactivity and Activity Series
To determine the outcome of the reaction \(Na_2CO_3 + K \rightarrow ?\), we must analyze the chemical reactivity of the metals involved: sodium (\(Na\)) and potassium (\(K\)). The activity series acts as a guide to predict whether a displacement reaction will occur. In this reaction, potassium is being added to a compound of sodium. A fundamental principle in chemistry is that a more reactive metal will displace a less reactive metal from its compound. By examining the provided activity series, we identify the positions of both metals. The series begins with \(Li > K > Ba > Sr > Ca > Na > ...\). We see that potassium (\(K\)) is listed before sodium (\(Na\)), which indicates that potassium is more reactive than sodium.
- Predicting the Products and Conclusion
Because potassium is more reactive than sodium, it has the chemical potential to displace the sodium from the sodium carbonate (\(Na_2CO_3\)) compound. This is a classic single-displacement reaction. Since the potassium is replacing the sodium atoms, the potassium atoms will bond with the carbonate ion (\(CO_3^{2-}\)) to form potassium carbonate (\(K_2CO_3\)), and the sodium atoms will be displaced as pure elemental sodium (\(Na\)). Therefore, the reaction will proceed as follows: \(Na_2CO_3 + K \rightarrow K_2CO_3 + Na\). Comparing this to the given options, it matches the second choice provided.
Examples
Imagine you are in a lab and trying to recover pure metal samples from ore. If you have a compound of a metal like sodium, knowing the activity series allows you to choose a 'helper' metal like potassium, which is stronger (more reactive) and can 'kick' the sodium out of its compound. This is a fundamental concept used in extractive metallurgy, where stronger reducing agents are used to isolate specific metals from their salts or ores in industrial manufacturing.