Asked by Lauren

2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O­(g) at STP

How many moles of O2 are needed to react with 60.0g of C8H18 (octane)?

mole of 02 = 60/114 = 53 mols

2 moles of octane =25 mols of 02

0.53/2 x 25= 6.625

answer is 6.6 mols

How many litres of CO2 are produced?

16 x 0.53mol/2mol = 4.24 moles

4.24mol x 22.4 L = 94.976

ANSWER = 95.0 L of co2

Answered by DrBob222
60/0/114 = 52,6 and not 53
so 60.0/114 x (25/2) = 6.5789 mols. You're allowed 3 s.f. (60.0 has 3) so round to 6.58 mols O2 needed.
Answered by Lauren
How many litres of CO2 are produced?

16 x 0.53mol/2mol = 4.24 moles

4.24mol x 22.4 L = 94.976

ANSWER = 95.0 L of co2
Answered by DrBob222
Using 52.6 and not 53, that is
60.0/114 x (16/2) x 22.4 = 94.3157 which rounds to 94.3 L to 3 S.F.
Answered by Anonymous
C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) This time we use 40 moles of O2 to combust, how many moles of H2O form?
Answered by Anonymous
C8H18(l) + 25 O2(g) ----> 16 CO2(g) + 18 H2O(g) Suppose we burn 24 moles of C8H18, how many moles of CO2 form?
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