If heat of combustion of acetylene C2H2 is -1301.1 kJ/mol then :
a) Write balanced thermochemical equation of acetylene combustion
b) If 0.25 mol C2H2 reacted according to the previous equation , and what is the emitted energy ?
c) how many grams needed to give off 3900 kJ energy ?
3 answers
How much of this do you know how to do. Show your work and let me go through it.
I think it's so :
a) 2C2H2 + 5O2 => 4CO2 + 2H2O + 1301.1kJ
b) 0.25mol * 1301.1 kJ = 325.27 kJ energy released
c) 1301.1 kJ/1mol = 3900 kJ/ x mol = 2.99mol
then X g = mol * molar mass of C2H2
X g = 2.99mol * 26 g/mol = 77.74 g
a) 2C2H2 + 5O2 => 4CO2 + 2H2O + 1301.1kJ
b) 0.25mol * 1301.1 kJ = 325.27 kJ energy released
c) 1301.1 kJ/1mol = 3900 kJ/ x mol = 2.99mol
then X g = mol * molar mass of C2H2
X g = 2.99mol * 26 g/mol = 77.74 g
All of this looks good to me except for the last part. Since the number of significant digits is 4 (at least I'm calling 3900 as 4 because I assume it is 3900.) then that 2.99 is really 2.997 ( or if not 4 s.f. for 3900 then it would be 2.997 and round to 3) which makes that 77.7 come out slightly higher as 77.9 or (78.0 if you use 3 x 26).
2 answers