How many molecules of iodine are produced when 9.3×1026 molecules of chlorine gas react with lithium iodide? __Cl2 + __LiI à __LiCl + __ I2

Explanation:

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In this case, considering the balanced chemical reaction:

Cl₂ + 2LiI ⇒ 2LiCl + I₂

We can see there is a 1:1 mole ratio between the produced iodine and the used chlorine, thus, we infer that the number of molecules of iodine given those of chlorine turn out:

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9.3 × 10²⁶ molecules of I₂

Explanation:

Step 1: Given data

Step 2: Write the balanced single displacement reaction

Cl₂ + 2 LiI ⇒ 2 LiCl + I₂

Step 3: Calculate the molecules of I₂ produced from 9.3 × 10²⁶ molecules of Cl₂

According to the balanced equation, the molecular ratio of Cl₂ to I₂ is 1:1.

9.3 × 10²⁶ molecule Cl₂ × 1 molecule I₂/1 molecule Cl₂ = 9.3 × 10²⁶ molecule I₂