To determine the maximum mass of sodium obtained from the electrolysis of 206 g of molten sodium bromide (NaBr), we need to follow these steps:
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Calculate the Molar Mass of NaBr:
- Sodium (Na) has an atomic mass of 23 u.
- Bromine (Br) has an atomic mass of 80 u.
- Therefore, the molar mass of NaBr = 23 u + 80 u = 103 g/mol.
-
Determine the Moles of NaBr:
- Given mass of NaBr = 206 g.
- Moles of NaBr = mass / molar mass = 206 g / 103 g/mol = 2 moles.
-
Electrolysis Reaction:
- The electrolysis of NaBr results in the formation of sodium (Na) and bromine (Br2).
- The balanced equation for the electrolysis of NaBr is:
[
2 \text{NaBr} \rightarrow 2 \text{Na} + \text{Br}_2
]
- From the equation, 2 moles of NaBr produce 2 moles of Na.
-
Calculate the Mass of Sodium Produced:
- Moles of Na produced = 2 moles (from the stoichiometry of the reaction).
- Molar mass of Na = 23 g/mol.
- Mass of Na = moles × molar mass = 2 moles × 23 g/mol = 46 g.
Therefore, the maximum mass of sodium that can be obtained from the electrolysis of 206 g of NaBr is 46 g. This supports Option 2 as the correct answer.
Explanation for Incorrect Options:
-
Option 1 (40 g): This option is incorrect because it underestimates the mass of sodium produced. The stoichiometry and calculations clearly show that 46 g of sodium is produced, not 40 g.
-
Option 3 (52 g): This option is incorrect because it overestimates the mass of sodium produced. The calculations based on the balanced chemical equation and molar masses show that only 46 g of sodium can be obtained.
-
Option 4 (23 g): This option is incorrect because it represents the mass of sodium from only 1 mole of NaBr, not the 2 moles present in 206 g of NaBr. The correct calculation for 2 moles results in 46 g of sodium.
In conclusion, the correct answer is Option 2, which accurately reflects the maximum mass of sodium obtained from the electrolysis of 206 g of sodium bromide.
