Hi there, I need help with these two chemistry questions. Please show your work.

Question

Hi there,

I need help with these two chemistry questions. Please show your work. Thank you.

Q1) A 9.82 g sample of 2-ethylhexanal (formula C8H16O), is mixed with 29.39 atm of O2 in a 1.96 L combustion chamber at 333 °C. The combustion reaction to CO2 and H2O is initiated and the vessel is cooled back to 333 °C. Report all answers to two decimal places in standard notation (i.e. 1.23 atm).

A) What is PCO2 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. Answer in atm.

Last answer: 1.947 atm (wrong)

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

Q2) A block of an alloy of mass 40.0 g and at a temperature of 14.51 °C is placed in a calorimeter containing 47.0 g of dichlorobenzene at 88.81 °C. If the final temperature of the alloy and dichlorobenzene is 60.56 °C, calculate the specific heat (in J/g/K) of the metal. The specific heat of dichlorobenzene is 1.13 J/g/K. Express your answer to three significant figures in scientific notation.

Last answer: 1.52 J/g/K. (Wrong)

Explanation / Answer

Balance the overall reaction first:

C8H16O + 23/2O2 ---------> 8CO2 + 8H2O

molecular Weight of C8H16O is 128 g/mol and the O2 is 32 g/mol so:

moles of C8H16O = 9.82 g / 128 g/mol = 0.0767 moles

moles of O2 = 29.39 * 1.96 / (0.082 * 333+273) = 1.16 moles

so if 1 mol of C8H16O -----> 23/2 mol O2 then 0.0767 will react with:

0.0767 * 23/2 = 0.8821 moles and we have 1.16 so the Oxygen is in excess and the limitant reactant is the C8H16O.

1 mole of C8H16O produce 8 moles of CO2 so:

0.0767 * 8 = 0.6136 moles of CO2

Assuming the same temperature and volume, the pressure of CO2 would be:

pCO2 = 0.6136 * 0.082 * (333+273) / 1.96 = 15.56 atm

For part b) the remanent moles of O2 are:

1.16 - 0.8821 = 0.2779 moles

PO2 = 0.2779 * 0.082 * 606 / 1.96

PO2 = 7.05 atm

Post the second question in a different question post, and I'll gladly help you there too. Now you should get the right answer. Hope this helps

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