If 40 grams of pentane are burned, how many grams of oxygen are needed ?

1) Molar mass C5H12= 5*12 +1 *12=60+12=72 g/mol
2) 40g C5H12 * 1 mol C5H12/72 g C5H12 = 40/72 mol C5H12
 3)                              C5H12 + 8O2 > 5CO2 + 6H2O
by reaction                 1 mol      8 mol
from problem             40/72 mol  x mol

x=(40/72) * 8/1=(40*8)/72=(40)/9 mol O2
4) M(O2)=2*16 g/mol =32 g/mol
5) (40)/9 mol O2 *(32 g O2/ 1 mol )=(40 * 32)/9 =142.2 g O2