The combustion of octane in excess air can be represented as
C8H18 (g) + 12 1/2O2⟶ 8CO2 (g) + 9H2O (g)
To produce 176 g of CO2 in his reaction
A. 1.0 mol of C8H18 needs to react with 25 mol of O2
B. 0.500 mol of C8H18 needs to react with 6.25 mol of O2
C. 0.500 mol of C8H18 needs to react with 3.125 mol of O2
D. 1.0 mol of C8H18 needs to react with 3.125 mol of O2
EDIT: okay! So I think I figured it out... I came up with the answer of B, (see attachment for my working out
) please let me know if you think I’ve gone wrong somewhere. Thank you!
Ugghhh I’m using an iPad and I’ve been trying to take a photo then shrink it using an app but my image still won’t get accepted and I’m not too sure what to do...
Anyone else use an iPad or a phone to do this stuff???