Consider the following unbalanced reaction: C8H18(l) +O2(g)-------->H2O (l) + CO
ID: 1038965 • Letter: C
Question
Consider the following unbalanced reaction:
C8H18(l) +O2(g)-------->H2O (l) + CO2(g)
a) What is the theoretical volume of water generated (in L) if 24.5 L of C8H18(density = 0.703 g/mL) were reacted with an infinite amount of oxygen?
b)If 1000. mL of C8H18 and 100. g of oxygen gas were combined in a reactor, what would be the limiting reagent?
c) If 375 mL of water were generated after reacting 24.5 L of C8H18 with an infinite amount of oxygen, what is the overall yield for the reaction?
d) What is the minimum amount of O2 necessary to produce 1000. kg of CO2?
Explanation / Answer
a) 2C8H18(l) + 25O2(g)--------> 18H2O (l) + 16CO2(g)
2 mol C8H18(l) = 25 mol O2
No of mol of c8H18 reacted = v*d/Mwt
= 24.5*10^3*0.703/114 = 151.1 mol
No of mol of water formed = 151.1*18/2 = 1360 mol
mass of water formed = n*Mwt = 1360*18 = 24480 g
volume of water formed = w/density = 24480/1 = 24480 ml = 24.48 L
b) No of mol of c8H18 reacted = v*d/Mwt
= 1000*0.703/114 = 6.167 mol
No of mol of O2 gas = w/Mwt = 100/32 = 3.125 mol
limiting reactant = O2
c) mass of water formed(practical yield) = v*d = 375*1 = 375 g
No of mol of c8H18 reacted = v*d/Mwt
= 24.5*10^3*0.703/114 = 151.1 mol
No of mol of water formed = 151.1*18/2 = 1360 mol
mass of water formed(theoretical yield) = n*Mwt = 1360*18 = 24480 g
%yield = P.Y/T.Y*100
= 375/24480*100
= 1.532%
C) NO of mol of Co2 produced = 1000/44 = 22.73 kmol
25 mol O2 = 16 mol CO2
No of mol of O2 required = 22.73*25/16 = 35.51 kmol
amounr of O2 O2 required = n*Mwt = 35.51*32
= 1136.32 kg