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How many molecules of iodine are produced when 9.3×1026 molecules of chlorine gas react with lithium iodide?

__Cl2 + __LiI à __LiCl + __ I2

Respuesta :

sebassandin

Answer:

[tex]9.3x10^{26}molec\ I_2[/tex]

Explanation:

Hello!

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:

[tex]9.3x10^{26}molec\ Cl_2*\frac{6.022x10^{23}molec\ I_2}{6.022x10^{23}molec\ Cl_2} =9.3x10^{26}molec\ I_2[/tex]

Best regards!

dsdrajlin

Answer:

9.3 × 10²⁶ molecules of I₂

Explanation:

Step 1: Given data

  • Molecules of I₂: ?
  • Molecules of Cl₂: 9.3 × 10²⁶ molecules

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