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CHEM - Faraday and Mole Calculation Example 2

Chemistry, Faraday, mole calculations- IBID | DSE | GCE | IAL | AP Chemistry

· chemistry,question analysis,mole calculations,Faraday

Here is another example of mole calculation🧐

**Step 1:**

Start by writing down the equation

**2Cl ^{-} > Cl_{2} + 2e^{- }**

**Step 2:**

Find the number of moles in the 18 cm^{3} of chlorine gas

**Volume = Moles × Molar Volume**

❗️❗️ Remember to use the same units (cm^{3})

By rearranging the above equation,

**Moles = 18 cm ^{3 }/ 24000 cm^{3 } = 0.00075 moles**

**Step 3:**

Note that electricity always refers to the flow of electrons

From the half equation, 2 electrons are needed to produce chlorine gas

**Hence, 2e ^{- }= 96500 × 2 = 193000 C**

**Final Step**💪

Don't forget about the moles!

**Multiply the number of coulombs by the amount of moles calculated earlier**

193000 × 0.00075 = 144.75 Coulombs

Therefore, 144.75 Coulombs are required to produce 18 cm^{3} of chlorine gas

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