I have a major chemistry test (lower college level...Chem 1) and am stuck on a f

Question

I have a major chemistry test (lower college level...Chem 1) and am stuck on a few problems. The answers I keep getting are not one of the multiple choice options. Help! Please explain the answers so I know what I am doing wrong!

1.) What volume of O2, measured at 225 degrees C and 0.970 atm, will be produced by the decomposition of 3.16 g KClO3? (R= 0.08206 L*atm/mol*K)

2 KClO3(s) ---> 2 KCl (s) + 3O2 (g)

a) 1.09 L

b) 1.24 L

c) 1.63 L

d) 3.26 L

e) 52.1 L

2.) If 46.1 g Cu at 11.6 degrees C is placed in 85.0 g H20 at 72.4 degrees C, what is the final temperature of the mixture? The specific heat capacities of copper and water are 0.385 J/g*k and 4.184 J/g*k, respectively.

a) 71.2 degrees C

b) 63.6 degrees C

c) 51.0 degrees C

d) 42.0 degrees C

e) 69.5 degrees C

3.) What is the minimum mass of ice at 0.0 degrees C that must be added at 1.00 kg of water to cool the water from 28.0 degrees C to 12.0 degrees C? (Heat of fusion = 333 J/g; specific heat capacities; ice = 2.06 J/g*k, liquid water = 4.184 J/g*K)

a) 175 g

b) 201 g

c) 244 g

d) 299 g

e) 1140 g

Explanation / Answer

(1) The answer is: c) 1.63 L

2 KClO3 => 2 KCl + 3 O2

Moles of KClO3 = mass/molar mass of KClO3

= 3.16/122.55 = 0.025785 mol

Moles of O2 = 3/2 x moles of KClO3

= 3/2 x 0.025785 = 0.038678 mol

Ideal gas equation: PV = nRT

Volume V = nRT/P

= 0.038678 x 0.08206 x (273.15 + 225)/0.970 = 1.63 L

(2) The answer is: e) 69.5 degrees C

Let T be the final temperature

Heat gained by Cu = mass x specific heat x temperature change of Cu

= 46.1 x 0.385 x (T - 11.6)

Heat lost by water = mass x specific heat x temperature change of water

= 85.0 x 4.184 x (72.4 - T)

Heat gained = heat lost

46.1 x 0.385 x (T - 11.6) = 85.0 x 4.184 x (72.4 - T)

373.3885T = 25954.2186

Final temperature T = 69.5 deg C

(3) The answer is: a) 175 g

Let m be the mass of ice

Heat gained by ice = mass x heat of fusion + mass x specific heat x temperature change of melted ice (water)

= m x 333 + m x 4.184 x (12.0 - 0) = 383.208m J

Heat lost by water = mass x specific heat x temperature change of water

= 1000 x 4.184 x (28.0 - 12.0) = 66944 J

Heat gained = heat lost

383.208m = 66944

Mass of ice m = 175 g

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