The same amount of electricity was passed through molten cryolite containing `Al_(2)O_(3)` and NaCl. If 1.8 g of Al were liberated in one cell, the amount of Na liberated in the other cell is

Options

A. 4.6 g
B. 2.3g
C. 6.4 g
D. 3.2 g

Answer

To solve the problem, we need to determine the amount of sodium (Na) liberated when the same amount of electricity is passed through molten cryolite containing Al2O3 and NaCl, given that 1.8 g of aluminum (Al) is liberated in one cell.

### Step-by-Step Solution:

1. **Determine the moles of Aluminum (Al) produced:**
- The molar mass of aluminum (Al) is 27 g/mol.
- Therefore, the number of moles of aluminum produced can be calculated as:
\[ \text{Moles of Al} = \frac{\text{Mass of Al}}{\text{Molar mass of Al}} = \frac{1.8 \text{ g}}{27 \text{ g/mol}} = 0.06667 \text{ moles} \]

2. **Calculate the amount of electricity used:**
- From the electrochemical reaction, we know that 3 moles of electrons (3 Faraday) are required to produce 1 mole of Al (Al3+ + 3e- → Al).
- Therefore, the amount of electricity (in Faraday) used to produce 0.06667 moles of Al is:
\[ \text{Electricity (Faraday)} = 0.06667 \text{ moles} \times 3 = 0.20001 \text{ Faraday} \]

3. **Determine the moles of Sodium (Na) produced:**
- The reaction for sodium is Na+ + e- → Na. This means that 1 mole of Na is produced per mole of electrons.
- Since the same amount of electricity is passed through NaCl, the moles of sodium produced will be equal to the moles of electrons used:
\[ \text{Moles of Na} = 0.20001 \text{ moles} \]

4. **Calculate the mass of Sodium (Na) produced:**
- The molar mass of sodium (Na) is 23 g/mol.
- Therefore, the mass of sodium produced can be calculated as:
\[ \text{Mass of Na} = \text{Moles of Na} \times \text{Molar mass of Na} = 0.20001 \text{ moles} \times 23 \text{ g/mol} = 4.6 \text{ g} \]

### Final Answer:
The amount of sodium liberated in the other cell is **4.6 g**.

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