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bulgar [2K]
3 years ago
14

A student performs the reduction of 4-nitrobenzaldehyde (151.12 g/mol) with sodium borohydride (37.83 g/mol) in the presence of

ethanol. The student mixes 4.13 g of 4-nitrobenzaldehyde with 0.700 g of sodium borohydride and obtains 1.85 g of 4-nitrobenzyl alcohol (153.14 g/mol). What is the percent yield of this reaction
Chemistry
1 answer:
Leviafan [203]3 years ago
8 0

Answer:

65.4%

Explanation:

The redox reaction is a 1:1:1 reaction because the reagents suffer a double displacement reaction, and the substance that is substituted have the same charge (H+ and Br-), thus, we first need to know which of the reagents is the limiting.

Let's test the 4-nitrobenzaldehyde as the limiting. The mass needed for sodium borohydride (m) is the mass given of 4-nitrobenzaldehyde multiplied by the stoichiometric mass of sodium borohydride divided by the stoichiometric mass of 4-nitrobenzaldehyde. The stoichiometric mass is the number of moles in the stoichiometric representation (1:1:1) multiplied by the molar mass, so:

m = (4.13 * 37.83*1)/(151.12*1)

m = 1.034 g

So, the mass needed of the other reagent is larger than the mass that was given, so, it will be the limiting, and the stoichiometric calculus must be done with it.

The mass of the product that was expected is then:

m = (0.700*153.14*1)/(37.83*1)

m = 2.83 g

The percent yield is the mass that was formed divided by the expected mass, and then multiplied by 100%:

%yield = (1.85/2.83)*100%

%yield = 65.4%

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Conduct metric Titration of H_2(SO_4) and Ba(OH)_2 Write an equation (including states of matter) for the reaction between H_2(S
meriva

Answer:

a) H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)

b) H₂SO₄, H⁺, HSO₄⁻, SO₄²⁻. H₂O, H⁺, OH⁻.

c) H⁺, HSO₄⁻, SO₄²⁻

d) As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) At the equivalence point equivalent amounts of H₂SO₄ and Ba(OH)₂ react. The conducting species are Ba²⁺, SO₄²⁻, H⁺ and OH⁻.

f) After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

Explanation:

a) Write an equation (including states of matter) for the reaction between H₂SO₄ and Ba(OH)₂.

The <em>balanced equation</em> is:

H₂SO₄ + Ba(OH)₂ ⇄ BaSO₄(s) + 2 H₂O(l)   [1]

b) At the very start of the titration, before any titrant has been added to the beaker, what is present in the solution?

In the beginning there is H₂SO₄ and the ions that come from its <em>dissociation reactions</em>: H⁺, HSO₄⁻, SO₄²⁻. There is also H₂O and a very small amount of H⁺ and OH⁻ coming from its <em>ionization</em>.

H₂SO₄(aq) ⇄ H⁺(aq) + HSO₄⁻(aq)

HSO₄⁻(aq) ⇄ H⁺(aq) + SO₄²⁻(aq)

H₂O(l)  ⇄ H⁺(aq) + OH⁻(aq)

c) What is the conducting species in this initial solution?

The main responsible for conductivity are the <em>ions</em> coming from H₂SO₄: H⁺, HSO₄⁻, SO₄²⁻.

d) Describe what happens as titrant is added to the beaker. Why does the conductivity of the solution decrease? What is the identity of the solid formed? What is the conducting species present in the beaker?

As the titration takes place, reaction [1] proceeds to the right. The conductivity of the solution decreases because the amount of H⁺, HSO₄⁻, SO₄²⁻ decreases. The formed solid is barium sulfate BaSO₄. Since BaSO₄ is very insoluble, the main responsible for conductivity are still H⁺, HSO₄⁻ and SO₄²⁻,

e) What happens when the conductivity value reaches its minimum value (which is designated as the equivalence point for this type of titration)? What is the conducting species in the beaker?

At the <em>equivalence point</em> equivalent amounts of H₂SO₄ and Ba(OH)₂ react. Only BaSO₄ and H₂O are present, and since they are <em>weak electrolytes</em>, there is a small amount of ions to conduct electricity. The conducting species are Ba²⁺ and SO₄²⁻ coming from BaSO₄ and H⁺ and OH⁻ coming from H₂O.

f) Describe what happens at additional titrant is added past the equivalence point. Why does the conductivity of the solution increase? What is the conducting species present in the beaker?

After the equivalence point there is an excess of Ba(OH)₂. The ions Ba²⁺ and OH⁻ are responsible for the increase in the conductivity, being the major conducting species.

7 0
3 years ago
. How many grams of water would require 4400 joules
Alja [10]

Answer:

The specific heat of water is 4.18 J/g C.

Explanation:

q

=

m

C

s

Δ

T

Never forget that!

2200

=

m

⋅

4.18

J

g

⋅

°

C

⋅

66

°

C

∴

m

≈

8.0

g

5 0
2 years ago
PLS HELP!! it’s science how do waves travel through gas
HACTEHA [7]

By 'waves' do you mean sound waves? If so sound waves need to travel through things like solids, liquids and yup <u>gases. </u>When the waves travel they are vibrating the molecules in the matter. By doing this he molecules in solids are packed very tightly.

6 0
3 years ago
A tetraphenyl phosphonium chloride (TPPCl) powder (FW=342.39) is 94.0 percent pure. How many grams are needed to prepare 0.45 L
slega [8]

Answer:

5.41 g

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For tetraphenyl phosphonium chloride :

Molarity = 33.0 mM = 0.033 M (As, 1 mM = 0.001 M)

Volume = 0.45 L

Thus, moles of tetraphenyl phosphonium chloride :

Moles=0.033 \times {0.45}\ moles

Moles of TPPCl = 0.01485 moles

Molar mass of TPPCl = 342.39 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.01485\ g= \frac{Mass}{342.39\ g/mol}

Mass of TPPCl = 5.0845 g

Also,

TPPCl is 94.0 % pure.

It means that 94.0 g is present in 100 g of powder

5.0845 g is present in 5.41 g of the powder.

<u>Answer -  5.41 g</u>

5 0
3 years ago
Which noble gas is closer to magnesium
cricket20 [7]

I believe Neon is closest to Magnesium on the Periodic Table of elements.

Sorry if I'm wrong, hope I could help though! :)

3 0
2 years ago
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