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atroni [7]
3 years ago
10

For each of the following circumstances, indicate whether the calculated molarity of NaOH would be lower, higher or unaffected.

Explain your answer in each case. a. The inside of the pipet used to transfer the standard HCl solution was wet with water.b. you added 40 mL of water to the titration flask rather than 25ml. c. The buret, wet with water, was not rinsed with NaOH solution before filling the buret with NaOH solution. d. Five (5) drops of phenolphthalein were added to the solution to be titrated rather than three (3) drops.
Chemistry
1 answer:
irina [24]3 years ago
8 0

Answer:

a)calculated molarity of NaOH would be lower

b) calculated molarity of NaOH would be lower

c) calculated molarity of NaOH would be lower

d) calculated molarity of NaOH would be unaffected

Explanation:

Let us recall that the reaction of NaOH and HCl is as follows;

NaOH(aq) + HCl(aq) ----> NaCl(aq) + H2O(l)

Since the reaction is 1:1, when the number of moles of HCl reacting with NaOH is low due to dilution, the calculated molarity of NaOH also becomes less than it's accurate value.

When 40mL of water is added to the titration flask rather than 25ml of water, the acid is more dilute hence less number of moles of acid than necessary reacts with the base thereby yielding a less than accurate value of the molarity of NaOH.

If the burette wet with water is not rinsed with NaOH solution, the concentration of the NaOH in the burette decreases due to dilution with water and a less than accuracy value is calculated for the molarity of NaOH.

If five drops of phenolphthalein is used instead of one or two drops, there is no qualms since enough phenolphthalein may be added to ensure that a sharp end point is obtained.

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1.42 g H2 is allowed to react with 10.4 g N2 , producing 2.14 g NH3 . Part A What is the theoretical yield in grams for this rea
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Taking into account the reaction stoichiometry, the theorical yield for the reaction is 8.0467 grams of NH₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 H₂ + N₂ → 2 NH₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

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  • N₂: 1 mole
  • NH₃: 2 moles

The molar mass of the compounds is:

  • H₂: 2 g/mole
  • N₂: 28 g/mole
  • NH₃: 17 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • H₂: 3 moles ×2 g/mole= 6 grams
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  • NH₃: 2 moles ×17 g/mole= 34 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 28 grams of N₂ reacts with 6 grams of H₂, 10.4 grams of N₂ reacts with how much mass of H₂?

mass of H_{2} =\frac{10.4 grams of N_{2}x 6 grams of H_{2} }{28 grams of N_{2}}

<u><em>mass of H₂= 2.2286 grams</em></u>

But 2.2286 grams of H₂ are not available, 1.42 grams are available. Since you have less mass than you need to react with 10.4 grams of N₂, H₂ will be the limiting reagent.

<h3>Definition of theorical yield</h3>

The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

<h3>Theoretical yield in this case</h3>

Considering the limiting reagent, the following rule of three can be applied: if by reaction stoichiometry 6 grams of H₂ form 34 grams of NH₃, 1.42 grams of H₂ form how much mass of NH₃?

mass of NH_{3} =\frac{1.42 grams of H_{2} x 34 grams of NH_{3}}{6grams of H_{2} }

<u><em>mass of NH₃= 8.0467 grams</em></u>

Then, the theorical yield for the reaction is 8.0467 grams of NH₃.

Learn more about the reaction stoichiometry:

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