The partial pressure of methane in the mixture of methane and ethane has been 1 atm.
Partial pressure has been the pressure exerted by a gas in the solution or mixture. The partial pressure of each gas has been the total pressure of the gaseous mixture.
The partial pressure of the gas has been dependent on the volume, temperature, and concentration of the gas.
The given methane has a partial pressure of 1 atm in the 15 L vessel. The addition of ethane results in the change in the total pressure of the mixture, as there have been additional moles of solute that contributes to the solution pressure.
However, since there has been no change in the concentration and volume of methane, the pressure exerted by methane has been the same. Thus, the partial pressure of methane has been 1 atm.
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The mass percentage is 15.1465%.
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Answer:
0.641 moles of ethane
Explanation:
Based on the equation:
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)
We can determine ΔH of reaction using Hess's law. For this equation:
<em>Hess's law: ΔH products - ΔH reactants</em>
ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}
<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>
<em />
ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}
ΔH = -1559.7kJ/mol
That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:
1.00x10³kJ * (1mole ethane / 1559.7kJ) =
<h3>0.641 moles of ethane</h3>
