Hi again!!
a. The balanced equation is (the combustion in air means a reaction
with O2):
2 C2H6 + 7 O2 ---> 4 CO2 + 6 H2O
b. At STP one mol of gas has 22.4 L, then at STP 15 L will be
(15/22.4)= 0.67 moles, then we have 0.67 moles of ethane.
If the combustion of 2 moles of ethane produces 4 moles of CO2, the
combustion of 0.67 moles of ethane will produce (0.67*4/2) = 1.34
moles of CO2.
This 1.34 moles of CO2 will have (1.34*22.4 L) = 30 L
30 L of CO2 are formed by the combustion of 15 L of C2H6.
c. NOTE: You do not specify the conditions for this point, but I will
work using STP condition, if the condition is other, let me know by a
request of a clarification and I will gladly recalculate this problem
for you.
Again there are 0.67 moles of ethane reacting and for every 2 moles of
C2H6 burned, 6 moles of water vapor are produced, then for this 0.67
moles of ethane burned we will obtain (0.67*6/2) = 2 moles of H2O.
The molecular mass of the water is (2*1g + 16g) = 18g and this is the
weight of one mol in grams, then we will have (2*18g) = 36g of H2O
from the combustion of 15 L of C2H6.
Note: you can find the following balanced equation of the combustion
of ethane in air:
C2H6 + 3.5 O2 —-> 2 CO2 + 3 H2O
This reduction of a one mol of C2H6 is doing to simplify calculations
related to this combustion, wich is very important in industry for
example; see the following document from TeraSen Gas for reference:
"Natural Gas Characteristics and Combustion":
http://www.terasen.com/NR/rdonlyres/emc2qcljixzbh2e7c6qhjiytp6aaur5jcy7rgkllk2hl7q4inbsdwiuzearzv6wubi5fz6kspgimiualockqna4qrgb/NaturalGasCharacteristicsandCombustion.pdf
I hope this helps you. Please feel free for request for all the
clarifications needed.
Regards.
livioflores-ga |
,
0 Comments
)