The missing part of the question is shown in the image attached
Answer:
C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)
V= 70.4L of CO2
Explanation:
Equation of the reaction is:
C10H22(l) + 31/2 O2 (g)-----> 10CO2(g) + 11H2O(l)
Number of moles of decane = mass/ molar mass
Molar mass of decane= 122gmol-1
n= 0.370×10^3g/122gmol-1= 3.0 moles
T= 13°C +273=286K
P= 1atm
R= 0.082 atmLK-1mol-1
From :
PV= nRT
V= nRT/P
V= 3.0×0.082×286/1
V= 70.4L of CO2
