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alukav5142 [94]

3 years ago

10

A gas cylinder contains 2.0 mol of gas X and 6.0 mol of gas Y at a total pressure of 2.1 atm. What is the partial pressure of ga

s Y? Use StartFraction P subscript A over P subscript T EndFraction equals StartFraction n subscript a over n subscript T EndFraction..

2 answers:

vivado [14]3 years ago

4 0

Answer : The partial pressure of X and Y gases are, 0.525 and 1.575 atm respectively.

Explanation : Given,

Moles of X = 2.0 mole

Moles of Y = 6.0 mole

Total pressure = 2.1 atm

Now we have to calculate the mole fraction of X and Y.

$\text{Mole fraction of }X=\frac{\text{Moles of }X}{\text{Moles of }X+\text{Moles of }Y}$

$\text{Mole fraction of }X=\frac{2.0}{2.0+6.0}=0.25$

and,

$\text{Mole fraction of }Y=\frac{\text{Moles of }Y}{\text{Moles of }X+\text{Moles of }Y}$

$\text{Mole fraction of }Y=\frac{6.0}{2.0+6.0}=0.75$

Now we have to calculate the partial pressure of X and Y.

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 = 2.1 atm

$X_i$ = mole fraction of gas

$p_{X}=X_{(X)}\times p_T$

$p_{X}=0.25\times 2.1atm=0.525atm$

and,

$p_{Y}=X_{(Y)}\times p_T$

$p_{Y}=0.75\times 2.1atm=1.575atm$

Thus, the partial pressure of X and Y gases are, 0.525 and 1.575 atm respectively.

storchak [24]3 years ago

3 0

Answer:

1.6 is the answer

Explanation:

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The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

$\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}$

For the products, we have:

$\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}$

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$\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}$

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$\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}$

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$\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}$

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$\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}$

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$P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%$

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