Question

aktiv will undergo scheduled maintenance on oct 19 from 5am to 8am et. during this period, the platform may be temporarily unavailable. question 27 of 37 write a balanced chemical equation based on the following description: solid chromium reacts with solid iodine to produce solid chromium(iii) iodide. cr(s) + i₂(s) →

Explanation:

Step1: Write the un - balanced chemical equation

Chromium (\(Cr\)) reacts with iodine (\(I_2\)) to form chromium(III) iodide (\(CrI_3\)). The un - balanced equation is \(Cr(s)+I_2(s)\to CrI_3(s)\)

Step2: Balance the iodine atoms

On the left - hand side, the number of iodine atoms is \(2\) (from \(I_2\)), and on the right - hand side, it is \(3\) (from \(CrI_3\)). The least common multiple of \(2\) and \(3\) is \(6\). So, we put a coefficient of \(3\) in front of \(I_2\) and \(2\) in front of \(CrI_3\). The equation becomes \(Cr(s)+3I_2(s)\to 2CrI_3(s)\)

Step3: Balance the chromium atoms

After balancing iodine, we have \(2\) chromium atoms on the right - hand side. So, we put a coefficient of \(2\) in front of \(Cr\) on the left - hand side. The balanced equation is \(2Cr(s)+3I_2(s)\to 2CrI_3(s)\)

Answer:

\(2Cr(s)+3I_2(s)\to 2CrI_3(s)\)