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avanturin [10]
3 years ago
11

If 30g of HCl is reacted with excess NaOH, and 10g of NaCl is produced, what is the percent yield of the experiment?

Chemistry
1 answer:
omeli [17]3 years ago
6 0

Answer:

Percent yield = 20.85%

Explanation:

Given data:

Mass of HCl = 30 g

Mass of NaCl produced = 10 g

Percent yield of NaCl = ?

Solution:

Chemical equation:

HCl  + NaOH   →    NaCl  + H₂O

Number of moles of HCl:

Number of moles = mass/ molar mass

Number of moles = 30 g/ 36.5 g/mol

Number of moles = 0.82 mol

Now we will compare the moles of HCl with NaCl

                              HCl        :        NaCl

                                 1          :            1

                               0.82      :        0.82

Theoretical yield:

Mass of NaCl:

Mass = number of moles × molar mass

Mass = 0.82 mol × 58.5 g/mol

Mass = 47.97 g

Percent yield:

Percent yield = Actual yield / theoretical yield × 100

Percent yield = 10 g / 47.97 g × 100

Percent yield = 20.85%

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In the case of sodium chloride; Sodium Na has 1 electron in its outer orbital while Chlorine Cl has 7 electrons. Thus, Cl requires 1 electron to complete its octet. This electron is donated by Na.

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When a solution of 0.1 M Mg(NO3)2 was mixed with a limited amount of aqueous ammonia, a light white, wispy solid was observed, i
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<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of magnesium nitrate and aqueous ammonia (ammonium hydroxide) is given as:

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A white precipitate of magnesium hydroxide is formed in the above reaction.

Ionic form of the above equation follows:

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As, ammonium and nitrate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

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s344n2d4d5 [400]

Neutralization reactions are the reactions type which form salts.

Explanation:

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