Question

Question #2fd18

Answer 1

The amount of `CO_2` given out`rarr`83.21gms

Explanation:

Mass of 1 gallon of octane fuel`rArr`2.695gm

So mass of10 gallons of octane fuel`rArr`26.95gm

Moles of octane gas{10 gallons} as reactants:`(26.95)/(114)=`0.2364
`C_8H_18 + (25)/(2) O_2 rArr 8CO_2 + 9 H_2O`

1 Mole of `c_8H_18rArr8` Moles of `CO_2` under combustion reaction.

Therefore 0.2364 Moles of `C_8H_18rArr` 0.2364`xx`8 Moles of `CO_2`
`rArr`1.891mol

The amount of `CO_2`{in gms}=Molar mass`xx` no. of moles
`rArr` 44`(gm)/(mol)xx1.891mol=`83.21 gms