Final answer:
To find the mass of sodium hydroxide produced from 9.00 L of 0.85 M NaCl solution through electrolysis, we calculate the number of moles of NaCl and then use the molar mass of NaOH to find that 306 grams of NaOH are formed.
Step-by-step explanation:
Calculating the Mass of Sodium Hydroxide Formed
When electric current is passed through an aqueous solution of NaCl (common table salt), a chemical reaction occurs that results in the formation of sodium hydroxide (NaOH), chlorine gas (Cl₂), and hydrogen gas (H₂). This process is known as electrolysis. To calculate the mass of sodium hydroxide produced, we apply stoichiometry using the balanced chemical equation and the concentration of the initial NaCl solution.
The balanced chemical equation for the reaction is:
2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + H₂(g) + Cl₂(g)
According to this equation, 2 moles of NaCl produce 2 moles of NaOH. Given that we have 9.00 L of 0.85 M NaCl, we can calculate the number of moles of NaCl:
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- Moles of NaCl = Volume (in L) × Molarity (M) = 9.00 L × 0.85 M = 7.65 moles.
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- Based on the stoichiometry of the equation, 7.65 moles of NaCl will produce an equal amount of NaOH, hence 7.65 moles of NaOH are formed.
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- To find the mass of NaOH produced, we multiply the number of moles by the molar mass of NaOH (40.00 g/mol). Mass of NaOH = 7.65 moles × 40.00 g/mol = 306 g.
Therefore, the mass of sodium hydroxide formed from 9.00 L of 0.85 M NaCl solution is 306 grams.