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CHAPTER 3. CH7 PROBLEM SOLVING CLASS R.D. A. BOLINAS. http:// chempsclass.wordpress.com. 3.10 Calculate the molar mass of each of the following: (a) N 2 O 4 ( b ) C 8 H 10 ( c ) MgSO 4 ⦁7H 2 O ( d ) Ca(C 2 H 3 O 2 ) 2. a) N 2 O 4 = 2(14.01) + 4(16.00) = 92.02 g/ mol

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chapter 3

CHAPTER 3

CH7 PROBLEM SOLVING CLASS

R.D. A. BOLINAS

slide3
3.10 Calculate the molar mass of each of the following:(a) N2O4(b) C8H10(c) MgSO4⦁7H2O (d) Ca(C2H3O2)2
slide4

a) N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol

  • b) C8H10 = 8(12.01) + 10(1.008) = 106.16 g/mol
  • c) MgSO4•7H2O = 24.31 + 32.07 + 11(16.00) + 14(1.008) = 246.49 g/mol
  • d) Ca(C2H3O2)2 = 40.08 + 4(12.01) + 6(1.008) + 4(16.00) = 158.17 g/mol
slide5

3.14 Calculate each of the following quantities:(a) Total number of ions in 38.1 g of SrF2(b) Mass in kilograms of 3.58 mol of CuCl2 2H2O(c) Mass in milligrams of 2.88 x1022formula units of Bi(NO3)3⦁ 5H2O

slide7
3.18 Calculate each of the following:(a) Mass % of I in strontium periodate(b) Mass % of Mn in potassium permanganate
slide9

3.23 Hemoglobin, a protein in red blood cells, carries O2 from the lungs to the body’s cells. Iron (as ferrous ion, Fe2+) makes up 0.33 mass % of hemoglobin. If the molar mass of hemoglobin is 6.8x 104 g/mol, how many Fe2+ions are in one molecule?

slide11

3.26 What is the empirical formula and empirical formula massfor each of the following compounds?(a) C4H8(b) C3H6O3(c) P4O10(d) Ga2(SO4)3(e) Al2Br6

slide13

3.32 A 0.370-mol sample of a metal oxide (M2O3) weighs 55.4 g.(a) How many moles of O are in the sample?(b) How many grams of M are in the sample?(c) What element is represented by the symbol M?

slide15

3.38Write balanced equations for each of the following by inserting the correct coefficients in the blanks:(a) __Cu(NO3)2(aq) + __KOH(aq) ⟶__Cu(OH)2(s) + __KNO3(aq)(b) __BCl3(g) + __H2O(l) ⟶__H3BO3(s) + __HCl(g)(c) __CaSiO3(s) + __HF(g) ⟶__SiF4(g) + __CaF2(s) + __H2O(l)(d) __(CN)2(g) __H2O(l)⟶ __H2C2O4(aq) __NH3(g)

slide16

a) Cu(NO3)2(aq) + 2 KOH(aq) 

Cu(OH)2(s) + 2 KNO3(aq)

  • b) BCl3(g) + 3 H2O(l)  H3BO3(s) + 3 HCl(g)
  • c) CaSiO3(s) + 6 HF(g)  SiF4(g) + CaF2(s) + 3 H2O(l)
  • d) (CN)2(g) + 4 H2O(l)  H2C2O4(aq) + 2 NH3(g)
slide17

3.41 The molecular scene at right represents a mixture of A2 (blue) and B2(green) before they react to form AB3.(a) What is the limiting reactant?(b) How many molecules of product can form?

slide18

The reaction is A2 + 3 B2 → 2 AB3 so the mole ratio between A2 and B2 is 1:3.

  • a) With 3 A2 molecules present, 3 x 3 = 9 B2 molecules

would be required. Since you have only 6 B2 molecules, B2is the limiting reagent.

b) 6 B2 molecules x (2 AB3 / 3 B2) = 4 AB3 molecules

slide19

3.51 Calculate the maximum numbers of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g of water:Al2S3 + H2O ⟶ Al(OH)3 + H2S [unbalanced]What mass of the excess reactant remains?

slide21

3.66 Six different aqueous solutions (with solvent molecules omitted for clarity) are represented in the beakers below, and their total volumes are noted.

A B C

D E F

slide22

(a) Which solution has the highest molarity? (b) Which solutions have the same molarity? (c) If you mix solutions A and C, does the resulting solution have a higher, a lower, or the same molarity as solution B?

A B C

D E F

slide23

a) Solution B has the highest molarity as it has the largest number of particles, 12, in a volume of 50 mL.

  • b) Solutions A and F both have 8 particles in a volume of 50 mL and thus the same molarity. Solutions C, D, and E all have 4 particles in a volume of 50 mL and thus have the same molarity.
  • c) Mixing Solutions A and C results in 12 particles in a volume of 100 mL. That is a lower molarity than that of Solution B which has 12 particles in a volume of 50 mL or 24 particles in a volume of 100 mL.
slide24

(d) After 50. mL of water is added to solutionD, is its molarity higher, lower, or the same as that of solution F after 75 mL of water is added to it?

A B C

D E F

slide25

d) Adding 50 mL to Solution D would result in 4 particles in a total volume of 100 mL; adding 75 mL to Solution F would result in 4 particles in a volume of 100 mL. The molarity of each solution would be the same.

slide26
(e) How much solvent must be evaporated from solution E for it to have the same molarity as solution A?

A B C

D E F

slide27

e) Solution A has 8 particles in a volume of 50 mL while Solution E has the equivalent of 4 particles in a volume of 50 mL. The molarity of Solution E is half that of Solution A. Therefore half of the volume, 12.5 mL, of Solution E must be evaporated. When 12.5 mL of solvent is evaporated from Solution E, the result will be 2 particles in 12.5 mL or 8 particles in 50 mL as in Solution A.

slide28

3.69 Calculate each of the following quantities:(a) Volume in milliliters of 2.26 M potassium hydroxide thatcontains 8.42 g of solute(b) Number of Cu2 ions in 52 L of 2.3 M copper(II) chloride(c) Molarity of 275 mL of solution containing 135 mmol of glucose

slide30

H

3.83 One of the compounds used to increase the octane rating of gasoline is toluene (right).Suppose 20.0 mL of toluene(d = 0.867 g/mL) is consumedwhen a sample of gasoline burnsin air. (a) How many grams ofoxygen are needed for complete combustion of the toluene?(b) How many total moles of gaseous products form? (c) How many molecules of water vapor form?

C

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