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svp [43]
3 years ago
10

Potassium hydrogen phthalate is often used to standardize solutions of strong bases. Highly pure samples of this solid weak acid

are readily available. It is stable and easy to use in the laboratory. Calculate the molarity of a sodium hydroxide solution if 23.73 mL are required to titrate 0.5816 g of potassium hydrogen phthalate (C8H5O4K).
Chemistry
1 answer:
Mazyrski [523]3 years ago
3 0

Answer:

0.12 M

Explanation:

Molarity of sodium hydroxide = number of moles of sodium hydroxide / volume in liters

equation of reaction between NaOH and Potassium hydrogen phthalate

NaOH(aq) + KHC₈H₄O₄(aq)  KNaC₈H₄O₄(aq) + H₂O(aq)

molar mass of KHC₈H₄O₄ = 204.22g/mol

amount required to titrate with NaOH = 0.5816 g

mole of  KHC₈H₄O₄ = 0.5816 g / 204.22g/mol = 0.00285 mole

considering the equation, 1 mole of NaOH required 1 mole of KHC₈H₄O₄

0.00285 mole  KHC₈H₄O₄ will require 0.00285 mole of NaOH

Molarity = 0.00285 mole  / ( 23.73 ml / 1000ml) L = 0.12 M

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If you feed 100 kg of N2 gas and 100 kg of H2 gas into a
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Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of N_2 = 100 kg  = 100000 g

Mass of H_2 = 100 kg = 100000 g

Molar mass of N_2 = 28 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of N_2 and H_2.

\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of N_2 react with 3 mole of H_2

So, 3571.43 moles of N_2 react with 3571.43\times 3=10714.29 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 1 mole of N_2 react to give 2 mole of NH_3

So, 3571.43 moles of N_2 react to give 3571.43\times 2=7142.86 moles of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg

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Molar mass:

O2 = 16 x 2 = 32.0 g/mol              Mg = 24 g/mol

<span>2 Mg(s) + O2(g) --->2 MgO(s) 
</span>
2 x 24.0 g Mg -------------> 32 g O2
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