When a 6.79-g mixture of methane, CH4, and ethane, C2H6, is burned in oxygen at constant pressure, 369 kJ of heat is liberated. What is the percentage by mass of CH4 in the mixture? The standard enthalpies of combustion for CH4 and C2H6 are -890.3 kJ mol-1 and -1569.7 kJ mol-1, respectively.
Two equations and two unknowns.
Let X = g CH4
and Y = g C2H6
------------------
eqn 1 is X + Y = 6.79g
eqn 2 is heat generated by CH4 + heat generated by C2H6 = 369 kJ.
heat from CH4 is 890.3 kJ/mol x (X/16)
heat from C2H6 is 1569.7 kJ/mol x (Y/2*30) so all of that together is
X+Y = 6.79
(890.3)(X/16) + (1569.7)(Y/2*30) = 369
Solve for X and Y, then
%CH4 = (mass CH4/6.79)*100 = ?
%C2H6 = (mass C2H6/6.79)*100 = ?
To determine the percentage by mass of CH4 in the mixture, we need to calculate the moles of methane (CH4) and ethane (C2H6) present in the mixture.
Step 1: Calculate the moles of CH4 and C2H6:
To calculate the moles, we need to use the molar mass of each compound:
Molar mass of CH4 = 12.01 g/mol (C) + 4(1.01 g/mol) = 16.04 g/mol
Molar mass of C2H6 = 2(12.01 g/mol) + 6(1.01 g/mol) = 30.07 g/mol
Now we can calculate the moles using the given mass of the mixture:
Moles of CH4 = (mass of CH4 in the mixture) / (molar mass of CH4)
Moles of C2H6 = (mass of C2H6 in the mixture) / (molar mass of C2H6)
However, we don't know the individual masses of CH4 and C2H6 in the mixture, so we need to find them.
Step 2: Assume a total mass for the mixture:
Let's assume the total mass of the mixture is 100 g.
Step 3: Calculate the masses of CH4 and C2H6 in the mixture:
To calculate the masses, we can assume that the percent composition of CH4 is x%, and the rest (100-x)% is C2H6.
Mass of CH4 = (x/100) * 100 g
Mass of C2H6 = ((100 - x)/100) * 100 g
Step 4: Calculate the moles of CH4 and C2H6 using the assumed masses:
Moles of CH4 = (mass of CH4) / (molar mass of CH4)
Moles of C2H6 = (mass of C2H6) / (molar mass of C2H6)
Step 5: Use the combustion reactions and enthalpies to calculate heat released by CH4 and C2H6:
The balanced combustion reactions for CH4 and C2H6 are:
CH4 + 2O2 → CO2 + 2H2O
C2H6 + 7/2O2 → 2CO2 + 3H2O
The heat released by CH4 can be calculated using its standard enthalpy of combustion:
Heat released by CH4 = (moles of CH4) * (standard enthalpy of combustion of CH4)
The heat released by C2H6 can be calculated using its standard enthalpy of combustion:
Heat released by C2H6 = (moles of C2H6) * (standard enthalpy of combustion of C2H6)
The total heat released by the mixture will be the sum of the heat released by CH4 and C2H6:
Total heat released = Heat released by CH4 + Heat released by C2H6
Step 6: Set up the equation using the given total heat released and solve for x:
Total heat released = (x/100) * (moles of CH4) * (standard enthalpy of combustion of CH4) + ((100 - x)/100) * (moles of C2H6) * (standard enthalpy of combustion of C2H6)
Given: Total heat released = -369 kJ
Standard enthalpy of combustion of CH4 = -890.3 kJ/mol
Standard enthalpy of combustion of C2H6 = -1569.7 kJ/mol
Solve the equation for x to find the percentage of CH4 in the mixture.
For a more automated approach, you could use a software tool like a spreadsheet or a programming language to plug in the values and calculate the percentage.