Chem 1A Practice Exam III key

Chem 1A Practice Exam III



Answer the following questions. You must show your work to receive credit. You will be graded for units an
significant figures.

A) How much energy is released or absorbed when you convert 100.0 g of liquid water at 45.0 ºC to ice at –2
ºC? Specific heat: H2O(l) = (4.18 J/g ºC) H2O(g) = (2.03 J/g ºC) H2O(s) = (2.06 J/g ºC)
H2O heat of vaporization = 2260 J/g H2O heat of fusion = 333 J/g


(100.0 g)(4.18 J/g ºC )(0.0 – 45.0 ºC) = –18810 J

(100.0 g)( – 333 J/g) = – 33300 J

(100.0 g)(2.06 J/g ºC )(– 25.0 – 0.0 ºC) = – 5150 J

_________________________________________________________

Sum = – 57300 J




B) Given 90.0 g C2H6 and excess O2, what pressure of H2O, in mm of Hg, will you obtain in a 43.0 L flask at
230.0 ºC?

2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g)

PV = nRT, looking for pressure: P = nRT/V



n = (90.0 g C2H6 )(1 mol C 2H6 /30.0)(6 mol H2O/2 mol C2H6) = 9.00 moles of O 2

R = 0.08206 atm . L/mole . K

T = 230.0 + 273.15 = 503.15 K

P = (9.00 moles)( 0.08206)(503.15 K)/43.0 L = 8.64 atm

8.64 atm (760 mm Hg/1 amt) = 6570 mm of Hg