Question

sodium (na) and iodine gas (i₂) react to form sodium iodine (nai). which equation is a correct representation of the balanced equation? 2na + i₂ → 2nai nai → 2na + i₂ na → nai + i₂ i₂ → 2nai + 2na

Explanation:

Step1: Check the number of atoms on both sides

For the equation \(2Na + I_{2}\to2NaI\):

• Sodium (\(Na\)): On the left - hand side, there are \(2\) \(Na\) atoms. On the right - hand side, in \(2NaI\), there are \(2\) \(Na\) atoms.
• Iodine (\(I\)): On the left - hand side, in \(I_{2}\), there are \(2\) \(I\) atoms. On the right - hand side, in \(2NaI\), there are \(2\) \(I\) atoms.

Step2: Analyze other equations

• For \(NaI\to2Na + I_{2}\): This is a decomposition reaction (opposite of the formation of \(NaI\) from \(Na\) and \(I_{2}\)) and the number of atoms is not balanced for the formation reaction.
• For \(Na\to NaI+I_{2}\): There is no iodine on the left - hand side initially (violates the law of conservation of mass as iodine atoms appear out of nowhere for the formation of \(NaI\) and \(I_{2}\)).
• For \(I_{2}\to2NaI + 2Na\): There is no sodium on the left - hand side initially (violates the law of conservation of mass as sodium atoms appear out of nowhere for the formation of \(NaI\) and \(Na\)).

Answer:

\(2Na + I_{2}\to2NaI\)