Questions: Question Some single displacement reactions involve one halogen replacing a less reactive halogen. Cl2, replaces iodine in NaI , producing I2 and NaCl . Write the balanced single displacement reaction below. - Do not include the states of the reactants or products.

Question
Some single displacement reactions involve one halogen replacing a less reactive halogen. Cl2, replaces iodine in NaI , producing I2 and NaCl . Write the balanced single displacement reaction below.
- Do not include the states of the reactants or products.
Transcript text: Question Some single displacement reactions involve one halogen replacing a less reactive halogen. $\mathrm{Cl}_{2}$, replaces iodine in NaI , producing $\mathrm{I}_{2}$ and NaCl . Write the balanced single displacement reaction below. - Do not include the states of the reactants or products.
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Solution

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Solution Steps

Step 1: Identify the Reactants and Products

In the given single displacement reaction, chlorine (\(\mathrm{Cl}_2\)) is reacting with sodium iodide (\(\mathrm{NaI}\)) to produce iodine (\(\mathrm{I}_2\)) and sodium chloride (\(\mathrm{NaCl}\)).

Step 2: Write the Unbalanced Chemical Equation

The unbalanced chemical equation for the reaction is:

\[ \mathrm{Cl}_2 + \mathrm{NaI} \rightarrow \mathrm{I}_2 + \mathrm{NaCl} \]

Step 3: Balance the Chemical Equation

To balance the equation, ensure that the number of each type of atom is the same on both sides of the equation.

  • Chlorine (Cl): There are 2 chlorine atoms on the left side (\(\mathrm{Cl}_2\)), so we need 2 \(\mathrm{NaCl}\) on the right side.
  • Iodine (I): There are 2 iodine atoms on the right side (\(\mathrm{I}_2\)), so we need 2 \(\mathrm{NaI}\) on the left side.
  • Sodium (Na): With 2 \(\mathrm{NaI}\) on the left, we have 2 sodium atoms, which matches the 2 \(\mathrm{NaCl}\) on the right.

The balanced chemical equation is:

\[ \mathrm{Cl}_2 + 2\mathrm{NaI} \rightarrow \mathrm{I}_2 + 2\mathrm{NaCl} \]

Final Answer

The balanced single displacement reaction is:

\[ \boxed{\mathrm{Cl}_2 + 2\mathrm{NaI} \rightarrow \mathrm{I}_2 + 2\mathrm{NaCl}} \]

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