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# Exam1DKey - 1 Isooctane(C8H18 Mm = 114.2 g/mol is one of...

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1) Isooctane (C 8 H 18 , M m = 114.2 g/mol) is one of the principle components of gasoline. a) (16) Write a balanced equation for the combustion of isooctane, where the only products are carbon dioxide and water. 2 C 8 H 18 + 25 O 2 b 16 CO 2 + 18 H 2 O b) (22) Based on your equation in part a, how many liters of carbon dioxide, collected at 1 atm pressure and 25 o C, can be formed from the combustion of 1 gallon (2.62 kg) of isooctane 2620 g C 8 H 18 x 18 8 18 8 H C g 114.2 H C mol 1 x 18 8 2 H C mol 2 CO mol 16 = 183.54 mol CO 2 PV = nRT V = P nRT = atm) (1.00 K) (298 mol) - atm/K - L (0.0821 mol) (183.54 = 4490 L volume = 4490 L 2) 50.0 mL of 0.200 M Ca(OH) 2 are added to 40 mL of 0.100 M HCl, forming water and calcium chloride. a) (12) What is the limiting reactant in this reaction? Ca(OH) 2 : 25.0 mL x 0.100 mmol/mL = 10.0 mmol HCl : 40 mL x 0.100 mmol/mL = 4 mmol Need 2X HCl limiting Don’t have 2X HCl reactant HCl b) (24) Set up a reaction table for this reaction, where your entries are in millimoles (assume 0 millimoles of water and calcium chloride initially)

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## This note was uploaded on 10/03/2008 for the course CH 201 taught by Professor Warren during the Spring '07 term at N.C. State.

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Exam1DKey - 1 Isooctane(C8H18 Mm = 114.2 g/mol is one of...

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