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scoray [572]
2 years ago
14

Consider the sodium borohydride reduction of camphor to isoborneol. A reaction was performed in which 1.600 g of camphor was red

uced by an excess of sodium borohydride to make 1.062 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction. Theoretical yield: g Percent yield:
Chemistry
1 answer:
ki77a [65]2 years ago
6 0

Answer:

62.5%

Explanation:

Number of moles of camphor reacted= 1.6g/152.23 g/mol = 0.011 moles

Since the reaction is 1:1,  then 0.011 moles of isoborenol is also produced.

Theoretical yield of isoborenol produced = 154.25 g/mol * 0.011 moles = 1.698 g

% yield = actual yield/theoretical yield * 100

% yield =  1.062 g/1.698 g * 100

% yield =  62.5%

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What is the pH of 0.30 M ethanolamine, HOCH2CH2NH2, (Kb = 3.2 x 10−5)?
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Answer:

pH= 11.49

Explanation:

Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.

From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5

Using the formula below;

[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)

[OH^-] =√(3.2×10^-5 × 0.30M)

[OH^-]= √(9.6×10^-6)

[OH^-]=3.0984×10^-3

pOH= -log[OH^-]

pOH= -log 3.1×10^-3

pOH= 3-log 3.1

pH= 14-pOH

pH= 14-(3-log3.1)

pH= 11+log 3.1

pH= 11+ 0.4914

pH= 11.49

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3 years ago
Indicate the number of unpaired electrons for following: [noble gas]ns2np5
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The chemical equation shows iron(III) phosphate reacting with sodium sulfate. 2FePO4 + 3Na2SO4 Fe2(SO4)3 + 2Na3PO4 What is the t
slava [35]

<u>Answer:</u> The theoretical yield of iron(III) sulfate is 26.6 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of iron(III) phosphate = 20.00 g

Molar mass of iron(III) phosphate = 150.82 g/mol

Putting values in equation 1, we get:

\text{Moles of iron(III) phosphate}=\frac{20g}{150.82g/mol}=0.133mol

The given chemical equation follows:

2FePO_4+3Na_2SO_4\rightarrow Fe_2(SO_4)_3+2Na_3PO_4

As, sodium sulfate is present in excess. So, it is considered as an excess reagent.

Thus, iron(III) phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of iron(III) phosphate produces 1 mole of iron(III) sulfate

So, 0.133 moles of iron(III) phosphate will produce = \frac{1}{2}\times 0.133=0.0665moles of iron(III) sulfate

Now, calculating the mass of iron(III) sulfate from equation 1, we get:

Molar mass of iron(III) sulfate = 399.9 g/mol

Moles of iron(III) sulfate = 0.0665 moles

Putting values in equation 1, we get:

0.0665mol=\frac{\text{Mass of iron(III) sulfate}}{399.9g/mol}\\\\\text{Mass of iron(III) sulfate}=(0.0665mol\times 399.9g/mol)=26.6g

Hence, the theoretical yield of iron(III) sulfate is 26.6 grams

8 0
3 years ago
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