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3 years ago

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For the following reaction, 5.20 grams of propane (C3H8) are allowed to react with 22.5 grams of oxygen gas. propane (C3H8) (g)

oxygen (g) carbon dioxide (g) water (g) What is the maximum amount of carbon dioxide that can be formed

1 answer:

Sliva [168]3 years ago

4 0

Answer:

15.58g of CO₂ is the maximum amount that can be produced

Explanation:

The propane reacts with oxygen as follows:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

<em>Where 1 mole of propane reacts with 5 moles of oxygen</em>

To solve this question we need to find the moles of propane and oxygen to find limiting reactant using the chemical reaction:

<em>Moles propane -Molar mass: 44.1g/mol-:</em>

5.20g * (1mol / 44.1g) = 0.118 moles

<em>Moles oxygen -Molar mass: 32g/mol-:</em>

22.5g O₂ * (1mol / 32g) = 0.703 moles

For a complete reaction of 0.703 moles of oxygen are:

0.703 moles O₂ * (1mol C₃H₈ / 5mol O₂) = 0.141 moles of propane are necessaries. As there are just 0.118 moles of propane, <em>propane is limiting reactant.</em>

The moles of carbon dioxide that are produced are:

0.118 moles C₃H₈ * (3 moles CO₂ / 1 mol C₃H₈) = 0.354 moles CO₂

The maximum mass that can be produced is -Molar mass CO₂: 44.01g/mol-:

0.354 moles CO₂ * (44.01g / mol) =

15.58g of CO₂ is the maximum amount that can be produced

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Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

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Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

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Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

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$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

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$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

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Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

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