The combustion of octane (c8h18) is represented by the following equation:
2c8h18 + 25o2 > 16co2 + 18h20
how many grams of carbon dioxide (co2) are formed from 0.929 g of octane?

Molar mass :

C8H18 => 114.23g/mol
CO2 => 44.01 g/mol

2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

2 x 114.23 g 16 x 44.01 g
0.929 g ( mass of CO2)

mass of CO2 = 0.929 x 16 x 44.01 / 2 x 114.23

mass of CO2 = 654.16464 / 228.46

= 2.863 g of CO2

hope this helps!