Question #8ebae
1 Answer
Explanation:
I'm not given a reaction, but I'm going to assume that the ethane undergoes combustion, and thus the equation would be
What we can do first in solving this equation is using the given conditions to calculate the number of moles of ethane used up, using the idealgas equation:
Our known values are

#P = 200cancel("kPa")((1"atm")/(101.325cancel("kPa"))) = 1.97# #"atm"# 
#V = 16# #"L"# 
#R = 0.082057("L"·"atm")/("mol"·"K")# (universal gas constant) 
#T = 5000^"o""C" + 273 = 5273# #"K"#
I converted the units where necessary to make sure I had the units
Let's now plug in these variables, and solve for
Now, we can use this value and the coefficients in the chemical equation to find the relative number of moles of water vapor,
Now, let's again use the idealgas equation to solve for the volume of the
Known values:

#P = 1.97# #"atm"# 
#n = 0.219# #"mol"# 
#R = 0.082057("L"·"atm")/("mol"·"K")# 
#T = 5273# #"K"#
Plugging them in and solving for the volume,
Notice that the volume is simply triple that of ethane,
We therefore could also have solved this using just the volume and the coefficients: