Question

4. 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 mols

How many litres of CO2 are produced?

16 x 0.53mol/2mol = 4.24 moles

4.24mol x 22.4 L = 94.976 L of co2

Answers

DrBob222
Looks OK to me. I might suggest that you correctly rounded to 0.53 for mols O2 for the number of significant figures but you didn't follow that practice for the rest of the problem.

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