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Answer :
Final answer:
To neutralize 200 mL of 2.00 M HCl with 1.00 M NaOH, 0.400 L (400 mL) of NaOH is required. The mass of NaOH that reacted in this neutralization is 16.00 g. However, this mass is not listed in the given options, which suggests a possible error in the question or the provided choices.
Explanation:
To calculate the volume of 1.00 M aqueous sodium hydroxide (NaOH) neutralized by 200 mL of 2.00 M aqueous hydrochloric acid (HCl), we can use the stoichiometry of the reaction which is a one-to-one molar ratio:
- NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)
First, we need to calculate the number of moles of HCl we have:
- (2.00 mol/L) × (0.200 L) = 0.400 mol HCl
Since the reaction ratio is 1:1, we need the same number of moles of NaOH to react with HCl:
- 0.400 mol NaOH
And now we find the volume of 1.00 M NaOH needed:
- Volume = moles / concentration
- Volume = 0.400 mol / 1.00 mol/L = 0.400 L
To find the mass of sodium hydroxide produced, we use the molar mass of NaOH (approximately 40.00 g/mol):
- Mass = moles × molar mass
- Mass = 0.400 mol × 40.00 g/mol = 16.00 g
However, there might be a confusion here since the question also asks about the mass of sodium hydroxide "produced," which could be interpreted as asking for the mass of NaOH that reacted. Since we're dealing with the neutralization reaction, no additional NaOH is produced; instead, it reacts and forms water and NaCl. So, the actual interpretation should pertain to the initial mass that reacted. If the problem meant the mass that neutralized the acid, then the correct mass is 16.00 g, not listed in the options given.
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