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Anna35 [415]
2 years ago
9

If 3.00 liters of 2M HCl neutralizes 1.00 liter of NaOH, what is the molarity of the NaOH?

Chemistry
1 answer:
PilotLPTM [1.2K]2 years ago
7 0

Answer:

M_{base}=6M

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize this is a question about titration, which base solved by knowing that the HCl reacts with the NaOH in a 1:1 mole ratio, and therefore, we can write the following:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the molarity of the base, NaOH, as shown below:

M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} \\\\M_{base}=\frac{3L*2M}{1.00L}\\\\M_{base}=6M

Regards!

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<u>Answer:</u> The percentage yield of aspirin is 38.02 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

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Putting values in equation 1, we get:

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The chemical equation for the formation of aspirin follows:

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As, acetic anhydride is present in excess. So, it is considered as an excess reagent.

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By Stoichiometry of the reaction:

1 mole of salicylic acid produces 1 mole of aspirin.

So, 0.0730 moles of salicylic acid will produce = \frac{1}{1}\times 0.0730=0.0730mol of aspirin

Now, calculating the mass of aspirin from equation 1, we get:

Molar mass of aspirin = 180.16 g/mol

Moles of aspirin = 0.073 moles

Putting values in equation 1, we get:

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To calculate the percentage yield of aspirin, we use the equation:

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Putting values in above equation, we get:

\%\text{ yield of aspirin}=\frac{5.0g}{13.15g}\times 100\\\\\% \text{yield of aspirin}=38.02\%

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