Question

1. For the following reaction, calculate how many moles of NO₂ forms when 0.356 moles of the reactant completely reacts.

2 N2O5g→4 NO2g+O2g

2. For the following reaction, calculate how many moles of each product are formed when 0.356 moles of PbS completely react. Assume there is an excess of oxygen.

2PbSs+3O2g→2PbOs+2SO2g

3. For the following reaction, calculate how many grams of each product are formed when 4.05 g of water is used.

2 H2O→2 H2+O2

4. Determine the theoretical yield of P₂O₅, when 3.07 g of P reacts with 6.09 g of oxygen in the following chemical equation  

4 P+5O2→2P2O5

5. Determine the percent yield of the following reaction when 2.80 g of P reacts with excess oxygen. The actual yield of this reaction is determined to by 3.89 g of P₂O₅.

4 P+5O2→2P2O5

Instructions: Answer the following questions using the following information: ∆H_fus=6.02 kJ/mol; ∆H_vap= 40.7 kJ/mol; specific heat of water is 4.184 J/g∙˚C; specific heat of ice is 2.06 J/g∙˚C; specific heat of water vapor is 2.03 J/g∙˚C.

1. How much heat is required to vaporize 25 g of water at 100˚C?
2. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your answer to three significant figures)?
3. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no heat is lost to the surroundings, what is the final temperature of the entire water sample after all the ice is melted (report your answer to three significant figures)?

Acid/Base: The Water Constant and pH

Instructions: Determine if each solution is acidic, basic, or neutral.

1.[H3O+] = 1 x 10-10 M; [OH-] = 1 x 10-4 M

2. [H3O+] = 1 x 10-7 M; [OH-] = 1 x 10-7 M
3. [H3O+] = 1 x 10-1 M; [OH-] = 1 x 10-13 M
4. [H3O+] = 1 x 10-13 M; [OH-] = 1 x 10-1 M

Instructions: Calculate [OH-] given [H3O+] in each aqueous solution and classify the solution as acidic or basic.

5. [H3O+] = 2.6 x 10-3 M

6. [H3O+] = 2.6 x 10-8 M

7. [H3O+] = 1.6 x 10-2 M
8. [H3O+] = 4.3 x 10-3 M

Instructions: Calculate the [H3O+] of solutions a and b; calculate the [OH-] solutions c and d.

9. pH = 2.76

10. pH = 3.65

11. pOH = 3.65
12. pOH = 6.87   

Equilibrium and Le Chateleir's Principal

Instructions: Write the equilibrium expression for each chemical equation.

1. 2 H2Sg⇌2H2g+S2g
2. NH4HSs⇌NH3g+H2Sg

Oxidation-Reduction Reactions

Instructions: Assign an oxidation state to each element, ion, or molecule.

1. Cl^-   
2. Br₂
3. K  
4. Ca²⁺

Instructions: Assign an oxidation state to each atom in each compound.

5. NaCl
6. CH4
7. SbCl5
8. Al2O3

Instructions: Assign an oxidation state to each atom in each polyatomic ion.

9. NH4+
10. Cl04-
11. Cr2072-  
12. Mn04-

Instructions: Identify the oxidation state of P in each ion.

13. PO43-   
14. HPO42-
15. H2PO4-
16. H3PO4  

Answer 1

(1) 2N₂Os UNO2 + O2 - 2 mol N2O5 gives u mol NO2 0.356 mol Nels_will give = 4 x 0.356 -10.712 mold 2 (2) 2 Pb Sco) + 302 G 2260 + 2502 - 2 mol Pbs gives 2 roles of Pbo and 2 moles of Sor 0.356 mol phs will also give 0.356 moll of Pho and so2 2H₂ + O₂ (3) 2H20 - та : __M.W=___ 18 gmol' moles= may 0.225 mol (initial) Mw. -moles of affer = Reaction - 2gmos"!_32 g mol 0.225 mol 0.225 = 0.1125 mol May mol x M.w _ To.usg [3.6 g (4) up + 50₂ 2P₂O5 Mars = 3.07 g 6.ogg MW = 3ignol 32gmolt 142 227 groll moles = 0.099 0.19 - - 0.19 -0.099x5 0.099 x 2 4 4 = 0.06625 0.0495 mol . Mase of Pros = 0.0495 mol x 142 groll E 7.029 g

2 P₂O5 . (5) up + 50₂ Mass: 2.8g MW_319 mol initial Moles = 0.09 moles final = 142 goolt - 0.09_x2 0.045. moles = 142 groll mans of P2O5 (theoretical yield) 0.045 mol x = 6:39 g/ % yield = x 100 = 3.89 g actual yield Theoretical yield 100 6.39g 60.88 % (6) 1.39 mol --- qe mass of water = 259 moles = 25 g Temperature = 100°C 18 gmolt sMyap x moles 40.7.5J Molx 1.39 mol 56.53 kJ - - - a Difus (7) ice vapour 100e vapour los ice - water - - water - -4°C 0°C 0°C 100°c qemSoT __= 259 x 2.06 J/gºC x 4°C = 206 I Athus 6.02 kJ mort x 1.39 mol - < 3.36 kg - 8360 J 42 = 25g x 4.184 Ilgo x 100°C = 10460 I s Hvap = 56.53 kJ = 56530 J 2 = 25 g x 2.03 Jlgo x 5°C = 253.75 J - Heat required ( 2 _ : 206 + $360 + 10460 + 56530+ 253.75 = 75.809.75 I 275.81 kJ

shfus (8.32g) ice - → water water 0°C Dºc - Water (556) 25°C 2 water ToC 1 q2 = q + Dt fus Sesgx 4.184 Ilgºcx (T-25) = 8.329 x2-06 J /gºC x (T-01_1 + 8.329 x 6.02 17 mel? 18 gror e{-61- 26t5 11.13*2 + 2+82-6 87.4608 T 5397-6 - Es 61.71°C 230.12T - 5753_ 170139 I + 2792.6 I 8535.6_c40°C 212.98 basic Acid / Base & Water constant and pH if (120+7 [on-] thin and otherwise if (mot = con- then neutral 0 Neutral acidic 0 Basic ② Neutral ② Basic ③ Acidic ☺ COH) X (M30+) = 1x10-14 [on] = 1x 10-14 10-14 = 3.840 * 10-12 M 1 [70] 2.6*10-3 © Come) =_2014__ 3.846 X10-7M 2.6 X10-8 (1) [ch] - - - - 6 . 2S X10-13 M 7.6 X10-2

© Com-) = 10-14 - 2.326_x10-12 M. 4.3x10-3 - 3 2_10_2-762_1738_x10-3M ph - (10+) (120+) = -107 10-3-65 = D 2.239 x10-4M- ① ID [on] = com') = 10-poH = 10-6-87 - 10 - 3.65 = 2.239 x 10-4M D 1.385 X10-7 M. ④ K = CNHz] [ H25] 1 K = [12] ² [s27° (1257² - cí = +1 (16) Br₂ = 0 Nau (My Ca 2+ + 2 Shus 1H₂O₃ Nhut - A H= +1 No-3 Clou = C1 = +7 O=-2 Cr2O7 =) (r = +6 0= -2 MADE = Mn = +7 O = -2 & = = Na +1 c = -4 Sb=ts Al = +3 = -1 M7 + 1 (1) -1 0 = -2 - ® Dou? = xt (-2) x 4 = x + (-2)*4 = -3 -3 x = +5 I + x + (-2) x4 = -2 th pou = 2 + x + (-2) X Y = -1 #12p0y = _3_+_X + (-2)*4 = 0 In=ts - 19 = +5 x = 45