Question

The most important reaction that allows our cars to do work is octane (C5H18) combustion.

a) what are the initial and final oxidation states of carbon in this reaction?

2c8h18(l)+25o2(g)→16co2(g)+18h2o(g)

b) Theoretically speaking if we fully combust 3.0 kg of octane, and keep the products at STP, what volume of gas do we expect to be produced?

c) If the % yield for this combustion reaction is 90%, what is the actual volume?

d) Water at 0 degrees C has a vapor pressure of 4.6 mmHg; what is the partial pressure of the other product?

Answer 1

Ans

a) The initial oxiadtion number of carbon in octane will be -18/8 ~ -2.25

The final oxidation state of carbon in CO2 is +4.

b) 3.0 kg or 3000 g of octane = 3000 / 114.23 = 26.26 moles

2 moles of octane makes (16 + 18) = 34 moles of gas

So 26.26 moles of it will make ( 26.26 x 34) / 2 = 446.42 moles of gas

one mole of gas at STP occupies 22.4 L of volume

So 446.42 moles of gas will occupy 446.42 x 22.4

= 9999.8 L of volume

c) If the percent yield is 90% , then the actual volume will be :

(9999.8 x 90 ) / 100

= 8999.82 L

d) The vapour pressure of water at 0 degrees C is 4.6 mmHg

so the partila pressure of the other product will 0.