a.) Write the balanced complete combustion reaction for ethane (C2H6), producing CO2 and water vapour. Make sure to include phases. Also, make sure that the stoichiometric coefficients are whole numbers and are not fractions.
b.) What is the enthalpy of reaction (ΔH°rxn), in kJ, for the process in part (a)?
c.) Calculate the heat (q), in kJ, produced when 265.0 g of ethane burns completely.
d.) Write the balanced incomplete combustion reaction for ethane (C2H6), producing CO and water vapour. Make sure to include phases. Also, make sure that the stoichiometric coefficients are whole numbers and are not fractions.
e.) What is the enthalpy of the reaction (ΔH°rxn), in kJ, for the process in part (d)?
f.) Calculate the total heat (q), in kJ, produced by 265.0 g of ethane when 6.00% by mass of it burns incompletely and the remainder burns completely.
(...I tried so hard to do these questions, but I just can't do them! Please explain the steps, please!
2C2H6 + 7O2 ==> 4CO2 + 6H2O
b.
dHrxn = (n*dHf products) - (n*dHf reactants)
c.
q = dHrxn from b x (265g/2*molar mass C2H6)
d.e.f. Follow a,b,c steps.
Post your work if you get stuck.
What is the difference between writing a COMPLETE and an INCOMPLETE equations in a) and d)?
whats the chemical eqn for the combustion of ethane to form CO tho
Sure! I can help you with these questions step by step.
a.) To write the balanced complete combustion reaction for ethane (C2H6), we need to balance the number of atoms on both sides of the reaction. The reaction is as follows:
C2H6 + O2 --> CO2 + H2O
First, let's balance the carbons by putting the coefficient 2 in front of CO2:
C2H6 + O2 --> 2CO2 + H2O
Next, balance the hydrogens by putting the coefficient 3 in front of H2O:
C2H6 + O2 --> 2CO2 + 3H2O
Lastly, balance the oxygens by putting the coefficient 7/2 (which we can write as 3.5) in front of O2:
C2H6 + 7/2O2 --> 2CO2 + 3H2O
To avoid fractions in stoichiometric coefficients, we can multiply the entire equation by 2:
2C2H6 + 7O2 --> 4CO2 + 6H2O
Therefore, the balanced complete combustion reaction for ethane is:
2C2H6 + 7O2 --> 4CO2 + 6H2O
b.) To calculate the enthalpy of reaction (ΔH°rxn), we need the standard enthalpies of formation (ΔH°f) for each compound involved in the reaction. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states at a defined temperature and pressure.
Using the ΔH°f values:
ΔH°f(C2H6) = -84.68 kJ/mol
ΔH°f(CO2) = -393.51 kJ/mol
ΔH°f(H2O) = -241.82 kJ/mol
ΔH°rxn = ΣnΔH°f(products) - ΣnΔH°f(reactants)
ΔH°rxn = [(4)(-393.51 kJ/mol) + (6)(-241.82 kJ/mol)] - [(2)(-84.68 kJ/mol)]
Therefore, the enthalpy of reaction for the complete combustion of ethane is:
ΔH°rxn = -1558.8 kJ
c.) To calculate the heat (q) produced when 265.0 g of ethane burns completely, we can use the equation:
q = m * ΔH
where:
q = heat produced (in kJ)
m = mass of ethane (in grams)
ΔH = enthalpy of reaction (in kJ)
First, convert the mass of ethane to moles using its molar mass (30.07 g/mol):
moles of ethane = 265.0 g / 30.07 g/mol
Next, use the enthalpy of reaction calculated in part (b) to find the heat produced:
q = (moles of ethane) * (ΔH°rxn)
Plug in the values to calculate the heat produced.
d.) To write the balanced incomplete combustion reaction for ethane (C2H6), we need to balance the number of atoms on both sides of the reaction. The reaction is as follows:
C2H6 + O2 --> CO + H2O
First, let's balance the carbons by putting the coefficient 2 in front of CO:
C2H6 + O2 --> 2CO + H2O
Next, balance the hydrogens by putting the coefficient 3 in front of H2O:
C2H6 + O2 --> 2CO + 3H2O
Lastly, balance the oxygens by putting the coefficient 7/2 (which we can write as 3.5) in front of O2:
C2H6 + 3.5O2 --> 2CO + 3H2O
To avoid fractions in stoichiometric coefficients, we can multiply the entire equation by 2:
2C2H6 + 7O2 --> 4CO + 6H2O
Therefore, the balanced incomplete combustion reaction for ethane is:
2C2H6 + 7O2 --> 4CO + 6H2O
e.) The enthalpy of the reaction for the incomplete combustion of ethane can be calculated using the same method as in part (b). However, the enthalpies of formation for CO and CO2 are needed.
The standard enthalpy of formation for CO is -110.53 kJ/mol.
Using the ΔH°f values:
ΔH°f(CO) = -110.53 kJ/mol
ΔH°f(CO2) = -393.51 kJ/mol
ΔH°f(H2O) = -241.82 kJ/mol
ΔH°rxn = ΣnΔH°f(products) - ΣnΔH°f(reactants)
ΔH°rxn = [(4)(ΔH°f(CO)) + (6)(ΔH°f(H2O))] - [(2)(ΔH°f(C2H6)) + 7(ΔH°f(O2))]
Plug in the values and calculate the enthalpy of the reaction.
f.) To calculate the total heat produced when 6.00% by mass of ethane burns incompletely and the remainder burns completely, we need to find the mass of ethane that burns incompletely and the mass of ethane that burns completely.
First, calculate the mass of ethane that burns incompletely:
mass of ethane (incompletely burned) = 6.00% * 265.0 g
Next, calculate the mass of ethane that burns completely:
mass of ethane (completely burned) = 265.0 g - (mass of ethane (incompletely burned))
For the incompletely burned ethane, calculate the heat produced using the balanced incomplete combustion reaction and the values of ΔH°f for each compound involved.
For the completely burned ethane, calculate the heat produced using the balanced complete combustion reaction and the value of ΔH°rxn obtained in part (b).
Finally, calculate the total heat produced by adding the heat produced by the incompletely burned ethane to the heat produced by the completely burned ethane.
I hope this explanation helps you understand the steps to solve these questions! Let me know if you have any further questions.