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3 years ago

Calculate the percent by mass of C in pentaerythritol (C(CH2OH)4)

2 answers:

7 0

Formula for pentaerythritol = C5H12O4

mass of pentaerythritol = 5 (12) + 12 (1) + 4 (16)
= 60 + 12 + 64
= 138 amu

mass of C = 12 amu

% of C in C5H12O4 = [{5(12)}/ 138 ] × 100
= (60 / 138) × 100 = 43.47%

miskamm [114]3 years ago

7 0

Answer:

$C= 44.12%$ % of C

Explanation:

Hi, the empirical formula of the pentaerythritol is {tex]C_5H_{12}O_4[/tex]

The molecular weights are:

$M_C=12 g/mol$

$M_H=1 g/mol$

$M_O=16 g/mol$

Due to the empirical formula in 1 mol of pentaerythritol you have 5 mol of C, 12 mol of H and 4 mol O

Taking a calculation base of 1 mol:

$m_C=12 g/mol*5mol$

$m_C=60 g$

$m_H=1 g/mol*12mol$

$m_H=12 g$

$m_O=16 g/mol*4mol$

$m_O=64 g$

The total weight will be:

$m_{tot}=64 g +12 g +60 g= 136 gl$

The C%:

$C= \frac{m_C}{m_{tot}}*100%$

$C= \frac{60g}{136g}*100%$

$C= 44.12%$

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Explanation:

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3 years ago

A mixture of 75 mole% methane and 25 mole% hydrogen is burned with 25% excess air. Fractional conversions of 90% of the methane

son4ous [18]

Solution :

Consider a mixture of methane and hydrogen.

Take the basis as 100 moles of the mixture.

The mixture contains 75% of methane and 25% of hydrogen by mole and it is burned with 25% in excess air.

Moles of methane = 0.75 x 100

Moles of hydrogen = 0.25 x 100

The chemical reactions involved during the reaction are :

$CH_4+2O_2 \rightarrow CO_2 + 2H_2O$

$CH_4+1.5O_2 \rightarrow CO+2H_2O$

$H_2+0.5O_2 \rightarrow H_2O$

The fractional conversion of methane is 90%

Number of moles of methane burned during the reaction is = 0.9 x 75

= 67.5

Moles of methane leaving = initial moles of methane - moles of methane burned

= 75 - 67.5

= 7.5 moles

Fractional conversion of hydrogen is 85%

The number of moles of hydrogen burned during the reaction is = 0.85 x 25

= 21.25

Moles of hydrogen leaving = initial moles of hydrogen - moles of hydrogen burned

= 25 - 21.25

= 3.75 moles

Methane undergoing complete combustion is 95%.

$CO_2$ formed is = 0.95 x 67.5

= 64.125 moles

$CO$ formed is = 0.05 x 67.5

= 3.375 moles

Oxygen required for the reaction is as follows :

From reaction 1, 1 mole of the methane requires 2 moles of oxygen for the complete combustion.

Hence, oxygen required is = 2 x 75

= 150 moles

From reaction 3, 1 mole of the hydrogen requires 0.5 moles of oxygen for the complete combustion.

Hence, oxygen required is = 0.5 x 25

= 12.5 moles

Therefore, total oxygen is = 150 + 12.5 = 162.5 moles

Air is 25% excess.

SO, total oxygen supply = 162.5 x 1.25 = 203.125 moles

Amount of nitrogen = $203.125 \times \frac{0.79}{0.21}$

= 764.136 moles

Total oxygen consumed = oxygen consumed in reaction 1 + oxygen consumed in reaction 2 + oxygen consumed in reaction 3

Oxygen consumed in reaction 1 :

1 mole of methane requires 2 moles of oxygen for complete combustion

= 2 x 64.125

= 128.25 moles

1 mole of methane requires 1.5 moles of oxygen for partial combustion

= 1.5 x 3.375

= 5.0625 moles

From reaction 3, 1 mole of hydrogen requires 0.5 moles of oxygen

= 0.5 x 21.25

= 10.625 moles.

Total oxygen consumed = 128.25 + 5.0625 + 10.625

= 143.9375 moles

Total amount of steam = amount of steam in reaction 1 + amount of steam in reaction 2 + amount of steam in reaction 3

Amount of steam in reaction 1 = 2 x 64.125 = 128.25 moles

Amount of steam in reaction 2 = 2 x 3.375 = 6.75 moles

Amount of steam in reaction 3 = 21.25 moles

Total amount of steam = 128.25 + 6.75 + 21.25

= 156.25 moles

The composition of stack gases are as follows :

Number of moles of carbon dioxide = 64.125 moles

Number of moles of carbon dioxide = 3.375 moles

Number of moles of methane = 7.5 moles

Number of moles of steam = 156.25 moles

Number of moles of nitrogen = 764.136 moles

Number of moles of unused oxygen = 59.1875 moles

Number of moles of unused hydrogen = 3.75 moles

Total number of moles of stack gas

= 64.125+3.375+7.5+156.25+764.136+59.1875+3.75

= 1058.32 moles

Concentration of carbon monoxide in the stack gases is

$=\frac{3.375}{1058.32} \times 10^6$

= 3189 ppm

b). The amount of carbon monoxide in the stack gas can be decreased by increasing the amount of the excess air. As the amount of the excess air increases, the amount of the unused oxygen and nitrogen in the stack gases will increase and the concentration of CO will decrease in the stack gas.

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3 years ago

28) Consider a 21.0 mL sample of pure lemon juice with a citric acid (H3C6H5O7) concentration of 0.30M. a. How many moles of cir

Damm [24]

<h3>#a. Answer:</h3>

0.0063 mole

<h3>Solution and explanation:</h3>

We are given 21.0 mL citric acid with a concentration of 0.30 M

Part a requires we calculate the number of moles of citric acid.

We need to know how to calculate the concentration of a solution;

Concentration or molarity = Number of moles ÷ Volume of the solution

Thus;

Number of moles = Concentration × Volume

Hence;

Moles = 0.30 M × 0.021 L

= 0.0063 mole

<h3>#b. Answer</h3>

1.21 g citric acid

<h3>Solution</h3>

Part B

We are required to calculate the mass of citric acid in the sample

Number of moles of a compound is calculated by dividing its mass by its molar mass.

Molar mass of Citric acid = 192.124 g/mol

Moles of citric acid = 0.0063 mole

But; Mass = Number of moles × Molar mass

Mass of citric acid = 0.0063 mol × 192.124 g/mol

= 1.21 g citric acid

<h3>#c. Answer</h3>

4.167 mL

<h3>Solution:</h3>

Part C

We are required to determine the initial volume before dilution;

We have;

Initial concentration (M1) = 0.30 M

Final volume (V2) = 250 mL or 0.25 L

Final concentration (M2) = 0.0050 M

Using the dilution formula we can get the initial volume;

Therefore, since; M1V1 =M2V2

V1 = M2V2÷M1

= (0.0050 × 0.25)÷ 0.30

= 0.004167 L or

= 4.167 mL

Therefore, the initial volume of the solution is 4.167 mL

8 0

3 years ago

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Answer:

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Explanation:

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For example:

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Similarly, when highlinium-308 undergoes positron emission the new element burienium is produced and the atomic number will be 307 while atomic mass remain same.

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