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3 years ago

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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Answer is: the % ionization of hypochlorous acid is 0.14.

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HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq).

Ka = [H⁺] · [ClO⁻] / [HClO].

[H⁺] is equilibrium concentration of hydrogen cations or protons.

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3 years ago

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It’s B

Explanation:

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3 years ago

What are the three characteristics properties of an acid

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If the energy of the reactants is less than

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2 years ago

A 1.757-g sample of a / alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of

kondor19780726 [428]

Answer:

78.14% Pb²⁺ and 21.86% of Cd²⁺

Explanation:

The first titration involves the reaction of both Pb²⁺ and Cd²⁺

In the second titration, as the buffer is HCN/NaCN, the Cd²⁺ precipitates as Cd(CN)₂ and the only ion that reacts is Pb²⁺

In the first titration:

Moles EDTA = Moles Pb²⁺ and Cd²⁺:

28.89mL = 0.02889L * (0.06950moles / L) = 2.008x10⁻³ moles in the aliquot. In the sample:

2.008x10⁻³ moles * (250.0mL / 50.0mL) =

0.01004 moles = Pb²⁺ + Cd²⁺ (1)

In the second titration:

19.07mL = 0.01907L * (0.06950mol / L) = 1.325x10⁻³ moles Pb²⁺ in the aliquot. In the sample:

1.325x10⁻³ moles Pb²⁺ * (250.0mL / 50.0mL) =

6.626x10⁻³ moles Pb²⁺

That means the moles of Cd²⁺ are:

0.01004 moles = Cd²⁺ + 6.626x10⁻³ moles Cd²⁺

3.413x10⁻³ moles Cd²⁺

The mass of each ion is:

Cd²⁺ -Molar mass: 112.411g/mol-:

3.413x10⁻³ moles Cd²⁺ * (112.411g / mol) =

0.384g of Cd²⁺

Pb²⁺ -Molar mass: 207.2g/mol-:

6.626x10⁻³ moles Pb²⁺ * (207.2g / mol) =

1.373g of Pb²⁺

The percent mass of each ion is:

1.373g Pb²⁺ / 1.757g = 78.14% Pb²⁺

And:

0.384g of Cd²⁺ / 1.757g * 100 = 21.86% of Cd²⁺

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3 years ago