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

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%