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3 years ago

what is the concentration of an NaOH solution that requires 50 mL of a 1.25 M H2SO4 solution to neutralize 78.0 ml of NaOH

Chemistry

1 answer:

Dimas [21]3 years ago

3 0

Answer:

The answer is the concentration of an NaOH = 1.6 M

Explanation:

The most common way to solve this kind of problem is to use the formula

C₁ * V₁ = C₂ * V₂

In your problem,

For NaOH

C₁ =?? v₁= 78.0 mL = 0.078 L

For H₂SO₄

C₁ =1.25 M v₁= 50.0 mL = 0.05 L

but you must note that for the reaction of NaOH with H₂SO₄

2 mol of NaOH raect with 1 mol H₂SO₄

So, by applying in above formula

C₁ * V₁ = 2 * C₂ * V₂

(C₁ * 0.078 L) = (2* 1.25 M * 0.05 L)

C₁ = (2* 1.25 M * 0.05 L) / (0.078 L) = 1.6 M

So, the answer is the concentration of an NaOH = 1.6 M

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2 years ago

4.33 g of 3-hexanol were obtained from 5.84 g of hex-3-ene. Determine the percentage yield of 3-hexanol. a Determine the moles o

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Answer: The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

Explanation:

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

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

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol$

The chemical equation for the conversion of hex-3-ene to 3-hexanol follows:

$\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}$

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

So, 0.0711 moles of hex-3-ene will produce = $\frac{1}{1}\times 0.0711=0.0711mol$ of 3-hexanol

Now, calculating the mass of 3-hexanol from equation 1, we get:

Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

$0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g$

To calculate the percentage yield of 3-hexanol, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

$\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%$

Hence, the amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

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