Sure, let's analyze the question step by step to determine which displacement reaction is most likely to occur.
In a displacement reaction, a more reactive metal displaces a less reactive metal from its compound. To solve this problem correctly, we need to refer to the reactivity series of metals. The reactivity series lists metals in order of their ability to displace other metals, from most reactive to least reactive.
Here is a simplified reactivity series:
- Potassium (K)
- Sodium (Na)
- Calcium (Ca)
- Magnesium (Mg)
- Aluminum (Al)
- Zinc (Zn)
- Iron (Fe)
- Lead (Pb)
- Hydrogen (H)
- Copper (Cu)
- Mercury (Hg)
- Silver (Ag)
- Gold (Au)
Considering barium (Ba) is our target metal in these reactions, and we note where it stands in reactivity in comparison to other metals given in the options.
Given reactions:
A. [tex]\( BaCl_2 + Ca \rightarrow CaCl_2 + Ba \)[/tex]
B. [tex]\( BaCl_2 + Sr \rightarrow SrCl_2 + Ba \)[/tex]
C. [tex]\( BaCl_2 + 2Na \rightarrow 2NaCl + Ba \)[/tex]
D. [tex]\( BaCl_2 + Mg \rightarrow MgCl_2 + Ba \)[/tex]
Comparing the metals with barium:
- Calcium (Ca) is more reactive than barium.
- Strontium (Sr) is not significantly more reactive than barium.
- Sodium (Na) is more reactive than barium.
- Magnesium (Mg) is more reactive than barium.
To determine which reaction is most likely, we look at the metal that can displace barium most effectively. Sodium (Na) is higher up on the reactivity series compared to barium and other candidates in the options.
Thus, sodium (Na) should successfully displace barium (Ba) from barium chloride ([tex]\(BaCl_2\)[/tex]):
[tex]\[ BaCl_2 + 2Na \rightarrow 2NaCl + Ba \][/tex]
Hence, the correct answer is:
C. [tex]\( BaCl_2 + 2Na \rightarrow 2NaCl + Ba \)[/tex]
