- #1
fabrave21
- 2
- 0
2 question I don't get ... =\ any help would be appreciated.
I don't even know where to start ...
2. A sample of a hydrocarbon is combusted completely in O2 to produce 21.83 g CO2, 4.47 g H2O and 311 kJ of heat. A) What is the mass of the hydrocarbon sample that was combusted? B) What is the empirical formula of the hydrocarbon? C) Calculate the value of ΔHfº per empirical-formula unit of the hydrocarbon.7. Suppose you have some reaction: 2A ---> B and B ---> 2A The rate for the formation of B from A is 3.8 x 102 s-1 and the rate for the formation of A from B is 8.6 x 10-1 s-1. A) What is the equilibrium constant for the equilibrium reaction, 2A <----> B ?
B) Suppose that you started with 2 moles of A and at some later time you measure .9 moles of B in a total volume of 1L, is the reaction at equilibrium? Explain.
2. mols C = 21.83/44 =.496
mols H = 4.47/18.0 =,248 x2 =.496
mols = g/Mwt then mols x Mwt = g
The mass of C is .496 x12.0 = .5.97 g C
The mass of H = .496 x 1.00 = ,496 g H
Total mass is 6.47 g for the hydrocarbon
The empirical formula is the ratio of the mols of C and H which makes it C1H1
6.47/ 13 = .496 mols
heat of combustion per formula weight is 311KJ/ .498 mols
= 624.497992 kJ
7. ?
I don't even know where to start ...
Homework Statement
2. A sample of a hydrocarbon is combusted completely in O2 to produce 21.83 g CO2, 4.47 g H2O and 311 kJ of heat. A) What is the mass of the hydrocarbon sample that was combusted? B) What is the empirical formula of the hydrocarbon? C) Calculate the value of ΔHfº per empirical-formula unit of the hydrocarbon.7. Suppose you have some reaction: 2A ---> B and B ---> 2A The rate for the formation of B from A is 3.8 x 102 s-1 and the rate for the formation of A from B is 8.6 x 10-1 s-1. A) What is the equilibrium constant for the equilibrium reaction, 2A <----> B ?
B) Suppose that you started with 2 moles of A and at some later time you measure .9 moles of B in a total volume of 1L, is the reaction at equilibrium? Explain.
Homework Equations
The Attempt at a Solution
2. mols C = 21.83/44 =.496
mols H = 4.47/18.0 =,248 x2 =.496
mols = g/Mwt then mols x Mwt = g
The mass of C is .496 x12.0 = .5.97 g C
The mass of H = .496 x 1.00 = ,496 g H
Total mass is 6.47 g for the hydrocarbon
The empirical formula is the ratio of the mols of C and H which makes it C1H1
6.47/ 13 = .496 mols
heat of combustion per formula weight is 311KJ/ .498 mols
= 624.497992 kJ
7. ?
Last edited: