A 9.14 g sample of 3,7-dimethyl-2,6-octadien-1-ol (formula C10H18O), is mixed wi

Question

A 9.14 g sample of 3,7-dimethyl-2,6-octadien-1-ol (formula C10H18O), is mixed with 57.11 atm of O2 in a 6.60×10-1 L combustion chamber at 220.5 °C. The combustion reaction to CO2 and H2O is initiated and the vessel is cooled back to 220.5 °C. Report all answers to two decimal places in standard notation (i.e. 1.23 atm).

1. What is PCO2 (in atm) in the combustion chamber assuming the reaction goes to completion? Assume all water is water vapour and that the volume of the system does not change.

2. What is PO2 (in atm) in the combustion chamber after the reaction goes to completion?

Explanation / Answer

C10H18O+O2----> CO2+H20

First we need to balance the equation reaction

C10H18O+ 14O2---> 10CO2+H2O

The molecular mass of C10H18O is 154.25g/mol

So, 9.14g of C10H18O are equivalent to 9.14/154.25=0.06moles ofC10H18O

From equation reaction we can say that to burn 0.06moles of C10H18O are required 14*0.06=0.84 moles of O2

And the product of the reaction would 0.6moles of CO2 and 0.54moles of H20

Assuming that CO2 behave as an ideal gas

PV=nRT---> we can calculate P, replacing n=0.6 ; R=0.08205746; T=493.65K ; V=6.60*10-1L

PCO2=36.83atm

To calculate PO2 at the end of the process first we need to calculate the amount of O2 at the begining.

nO2=PV/RT with P=57,11atm; V=0.66L; R=0.08205746 and T=493.65K

nO2=0.93mol at the begining and 0.84moles of O2 were used in the reaction so 0.93-084=0.09 moles of O2 are left in the chamber

The PO2 can be calculate it with the ideal gas equation

PV=nRT with nO2=0.09; V=0.66L R=0.08205746 and T=493.65K

PO2 at the end=5.53atm

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