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3 years ago

What is the mass in grams of oxygen gas that is required to produce 32.1 grams of maleic acid anhydride?

1 answer:

3 0

Considering the reaction stoichiometry, the mass of oxygen gas that is required to produce 32.1 grams of maleic acid anhydride is 47.17 grams.

The balanced reaction is:

2 C₆H₆(l) + 9 O₂(g)→ 2 C₄H₂O₃(s) + 4 CO₂(g) + 4 H₂O(g)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

C₆H₆: 2 moles
O₂: 9 moles
C₄H₂O₃: 2 moles
CO₂: 4 moles
H₂O: 4 moles

The molar mass of each compound is:

C₆H₆: 78 g/mole
O₂: 32 g/mole
C₄H₂O₃: 98 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

By reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:

C₆H₆: 2 moles× 78 g/mole= 156 grams
O₂: 9 moles× 32 g/mole= 288 grams
C₄H₂O₃: 2 moles× 98 g/mole= 196 grams
CO₂: 4 moles× 44 g/mole= 176 grams
H₂O: 4 moles× 18 g/mole= 72 grams

Then you can apply the following rule of three: If by stoichiometries 196 grams of C₄H₂O₃ are produced by 288 grams of O₂, 32.1 grams of C₄H₂O₃ are produced from how much mass of O₂?

$mass of O_{2} =\frac{32.1 grams of C_{4} H_{2} O_{3} x288 grams of O_{2}}{196grams of C_{4} H_{2} O_{3}}$

mass of O₂= 47.17 grams

Finally, the mass of oxygen gas that is required to produce 32.1 grams of maleic acid anhydride is 47.17 grams.

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Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

Eduardwww [97]

Answer:

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$730.5mmHg\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 293}=0.3794mol$

According to the reaction shown below as:-

$2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)$

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when $\frac{2}{3}\times 0.3794$ moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate reacted = 0.2529 moles

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas = $20^oC=[20+273]K=293K$

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$735.5mmHg\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 293}=0.40211mol$

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol\times 32g/mol)=12.8675g$

Hence, the amount of oxygen gas collected will be 12.8675 g

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