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Chem 121 Problem set III Solutions - 1 Problem Set III Stoichiometry - Solutions

骣55.85 g Fe 1. mass Fe=( 2.25 mol Fe)琪 = 1.26 x 102 g Fe 桫1mol Fe

骣1mol Zn骣 6.022 x 1023 atoms Zn 2. Zn atoms=( 20.0 g Zn)琪 琪 = 1.84 x 1023 atoms Zn 桫65.38 g Zn桫 1mol Zn

3. molecular mass of ethane = 2(12.011) + 6(1.008) = 30.07 g 骣1molecule ethane molecules ethane=( 50.3 g ethane)琪 = 1.67 molecules 桫 30.07 g ethane

4. molecular mass of aniline = 6(12.011) + 7(1.008) + 14.01= 93.14 g/mol 骣6 x 12.01g C mass C=( 125.00 g C6 H 7 N)琪 = 96.71 g C 桫93.14 g C6 H 7 N

5. Let the molar mass of X be x.

1g1mol X4 O 10 10mol O 16.00g -3 mass of O in 40.0mg X4 O 10 =40.0mg创 创 = 22.5 x 10 mg 1000mg (4x+160)g 1mol X4 O 10 1mol O 40.0创 10 16.00 = 22.5 (4x+ 160) x = 31.1 g/mol and the element is P.

6. Let atomic mass of M be x

1mol MCl .2H O 2mol H O 18.016g mass of water formed from 0.642g hydrate =0.642g创 2 2 2 � 0.0949g (x+106.94)g 1mol MCl2 .2H 2 O 1mol H 2 O 0.642创 1 2 18.016 = 1.683 23.1325=0.0949x+10.1486 (x+106.94)创 1 1 solving for x = 137 g/mol which is Ba how would you check your answer?

7. Let atomic mass of M be x

4 M + 3 O2  2 M2O3 1mol M2mol M O (2x+ 48)g moles of oxide formed from 1.443g metal =1.443g创 2 3 � 1.683g x 4mol M 1mol M2 O 3 1.443创 1 2� (2x 48) = 1.683 5.772x+138.528=6.732x x创 4 1 solving for x = 144.3 g/mol which is Nd how would you check your answer?

8. Molecular mass of C2H5OH is (2x12.011 + 6x1.0079 + 15.999) = 46.069

2mol创 12.011g mol 6mol 1.0079g mol %C=� 100 = 52.144% %H � 100 13.127% %O=34.729% 46.069g mol 46.069g mol

Molecular mass of Ca(HCO3)2 is (40.078 + 2x1.0079 + 2x 12.011 + 6x15.999) = 162.11

%Ca = 24.72%, %C = 14.82%, %H = 1.24%, %O = 59.22%

Molecular mass of MgNH4PO4 is (24.305 + 14.007 + 4x1.0079 + 30.974 + 4x15.999) = 137.3 Chem 121 Problem set III Solutions - 2

Mg = 17.70%, N = 10.20%, H = 2.936%, P = 22.56%, O = 46.61%

9. - use CO2 to get mass of C and H2O to get mass of H

骣 骣 1mol CO2 1mol C骣 12.01g C mass C=( 2.23 g CO2 )琪 琪 琪 = 0.6086 g C 桫44.01g CO2 桫 1mol CO 2 桫 1mol C 骣 骣 1mol H2 O 2 mol H骣 1.008 g H mass H=( 1.37 g H2 O)琪 琪 琪 = 0.1533 g C 桫18.02 g H2 O 桫 1mol H 2 O桫 1mol H  quick check to see that C and H are the only components of this compound: 0.6086 g + 0.1533 g = 0.7619 g OK  if this added up to < 0.761 g, we would know that there must a 3rd element present, for which we could get the mass by difference 骣0.6086 g C 骣 0.1533 g C %C=琪( 100%) = 79.9 %C %H = 琪 ( 100%) = 20.1%H 桫0.761 g compound 桫 0.761g compound

10. The first step is to find the percentage composition of the unknown, as follows:

moles CO2  moles C in CO2  moles C in compound  mass C in compound  percent C

1 mole CO2 1 mole C 12.01 g C mass of C in Cx H y O z = 0.1486 g CO 2 x x x = 0.040555 g C 44.098 g CO2 1 mole CO 2 1 mole C 0.040555 g C percent C = x 100 = 40.0 % C 0.1014 g unknown moles H2O  moles H in H2O  moles H in compound  mass H in compound  percent H

1 mole H2 O 2 moles H 1.008 g H mass of H in Cx H y O z = 0.0609 g H 2 O x x x = 0.00681 g H 18.016 g H2 O 1 mole H 2 O 1 mole H 0.00681 g H percent H = x 100 = 6.72 % H 0.1014 g unknown Percentage of O in unknown = 100 - 40.0 - 6.72 = 53.3 %

The next step is to find the empirical formula from the percentage composition. I will assume I have 100.0 g of the unknown for convenience.

Element Relative mass Relative number of moles (atoms) Divide by the smallest number 40.0 3.33 = 3.33 = 1.0 C 40.0 12.01 3.33 6.72 6.67 = 6.67 = 2.0 H 6.72 1.008 3.33 53.3 3.33 = 3.33 = 1.0 O 53.3 16.00 3.33

The empirical formula is CxHyOz = C1H2O1 or CH2O

To find the molecular formula, the molecular mass needs to be known. Let's say that the molecular mass has been measured by mass spectrometry and found to be 180.0 amu. Since the empirical formula is CH2O, the empirical mass is 12.01 + 2 x 1.008 + 16.00 = 30.02 amu. Molecular mass 180.0 The ratio of molecular mass to empirical mass is = = 6.0 Empirical mass 30.02

So the molecular formula is (CH2O)6 or C6H12O6. Chem 121 Problem set III Solutions - 3

11. Take a 100.00 g sample

Atom Mass (g) Moles Divide by smallest Best ratio Na 32.79 32.79 / 22.99 = 1.4263 2.956 3 Al 13.02 13.02 / 26.982 = 0.48255 1 1 F 54.19 54.19 / 18.998 = 2.8523 5.911 6

Na3AlF6

1 mole H2 O 1 moles O 16.00g O 12. mass of O in Lax O y =0.05254g H 2 O创 x =0.046661g 18.016 g H2 O 1 mole H 2 O 1 mole O

mass of La in Lax O y = 0.3167 - 0.046661 = 0.27004

Element Relative mass Relative number of Divide by the smallest moles (atoms) number 0.27004 1.9441 10-3 = 1.9441 10-3 La 0.27004 -3 = 1.0 138.9 1.9441 10 -3 0.046661 -3 2.9163 10 O 0.046661 = 2.9163 10 -3 = 1.500 16.00 1.9441 10

The empirical formula is LaxOy = La2O3

13. you don’t have to come up with an equation but it can help

Ti + S  TixSy a) mass of product = 31.700 g – 11.120 g = 20.58 g product

100% of Ti used resulted in product formation

8.820 g Ti  the % of Ti in the product is: x 100%= 42.86 %Ti 20.58 g product and all the rest is S  100 – 42.86 = 57.14 %S

骣1mol Ti b. for the empirical formula, require moles (8.820 g Ti)琪 = 0.184 mol Ti 桫47.90 g Ti  sulphur was in excess,  use the % S in the product to get the amount of sulphur that must have reacted:

骣57.14 g S 骣 1mol S 琪(20.58 g product) 琪 = 0.3668 mol S 桫100 g product 桫 32.06 g S or could have got g S from 20.58 g – 8.820 g = 11.76 g S 骣1mol S and (11.76 g S)琪 = 0.3668 mol S 桫32.06 g S

Ti0.184/0.184 S0.367/0.184  Ti1S1.99 ~ TiS2 Chem 121 Problem set III Solutions - 4

14. Mo2O3  MoxOy 12.64g 13.48g The increase in mass is oxygen, so mass of extra O is 13.48 - 12.64 = 0.84 g

1mol Mo2 O 3 3mol O 16.00g but mass of O in Mo2 O 3 = 12.64g创 � 2.529g 239.88g 1mol Mo2 O 3 1mol O so total mass of oxygen in new oxide is 2.529g + 0.84g = 3.369g so mass of Mo in MoxOy = 13.48g - 3.369g = 10.1107g

Element Mass Relative number of Divide by the smallest Whole number moles (atoms) number ratio Mo 10.111 0.105386 1 1 O 3.369 0.21056 1.998 2

Empirical formula of the oxide is MoO2

15. N2O5 + H2O  2 HNO3

Mg2C3 + 4 H2O  2 Mg(OH)2 + C3H4

PCl5 + 4 H2O  H3PO4 + 5 HCl

16 Cr(s) + 3 S8(s)  8 Cr2S3(s)

Au2S3(s) + 3 H2(g)  2 Au(s) + 3 H2S(g)

6 NH4ClO4(s) + 10 Al(s) ® 3 N2(g) + 6 HCl(g) + 9 H2O(g)

16. - first write what you know from the question (you should know that bromine molecules, in fact all the halogens, exist as diatomic molecules)

Na + Br2  NaBr

- then balance it

2Na + Br2  2NaBr

17. - we know that when you combust a hydrocarbon, we get CO2 and H2O  and the oxidant in combustion is always oxygen (almost always):

C4H10 + O2  CO2 + H2O

13 - now balance it: C4H10 + /2O2  4CO2 + 5H2O - but we don’t want fractional coefficients:

2C4H10 + 13O2  8CO2 + 10H2O

18. - 1st we write what we know in an equation:

NH3 + CuO  N2 + Cu

- 2nd we balance it  but wait, we’ve only got H and O on the reacting side  there must be the formation of water in this reaction  can balance it now:

2NH3 + 3CuO  N2 + 3Cu + 3H2O

- can now do a  c 骣3 mol CuO a) mol CuO=( 0.445 mol NH3 )琪 = 0.668 mol CuO 桫2 mol NH3

骣1mol N2骣 28.02 g N 2 b) mass N2=( 3.18 mol CuO)琪 琪 = 29.7 g N 2 桫3 mol CuO桫 1mol N2

骣1mol Cu 骣2 mol NH3骣 17.04 g NH 3 b) mass NH3=( 55.0 g Cu)琪 琪 琪 = 9.83 g NH 3 桫63.55 g Cu 桫 3 mol Cu桫 1mol NH3 Chem 121 Problem set III Solutions - 5

1 mole WO3 1 mole W 183.9 g W 19. mass of W formed = 100.0 g WO3 x x x = 79.3 g W 231.9 g WO3 1 mole WO 3 1 mole W

1 mole WO3 3 mole H2 O 18.02 g H 2 O mass of H2 O formed = 100.0 g WO 3 x x x = 23.3 g H 2 O 231.9 g WO3 1 mole WO 3 1 mole H 2 O

1 mole WO3 3 mole H2 2.016 g H 2 mass of H2 used = 100.0 g WO 3 x x x = 2.61 g H 2 231.9 g WO3 1 mole WO 3 1 mole H 2 CHECK: The sum of the masses of reactants = the sum of the masses of products Reactants: 79.3 g + 23.3 g = 102.6 g Products: 100.0 g + 2.61 g = 102.6 g

20. 6 NH4ClO4 + 10 Al ® 5 Al2O3 + 3 N2 + 6 HCl + 9 H2O

31000g1mol NH4 ClO 4 10mol Al 26.98g Al 6 mass Al=5.0创 10 kg NH4 ClO 4 创 � 1.91 10 g 1kg 117.49g 6mol NH4 ClO 4 1mol Al

21. 2 C57H110O6 + 163 O2 ® 114 CO2 + 110 H2O

1000g 1mol fat110mol H O 18.02g mass water=1.0kg fat创 创 2 = 1111g = 1.1kg 1kg 891.5g 2mol fat 1mol H2 O

22. Let mass BaO2 = (x) g , thus mass BaCO3 = (14.53 - x)g Mass BaO formed:

1mol BaO2 1mol BaO 153.3g from BaO2= x g BaO 2 创 � 0.9055x g 169.3g 1mol BaO2 1mol BaO

1mol BaCO3 1mol BaO 153.3g from BaCO3= (14.53 - x)g BaCO 3 创 � - 0.77695(14.53 x) g 197.31g 1mol BaCO3 1mol BaO but 12.37g BaO formed=[0.9055x+0.77695(14.53-x)]g 0.9055x+ 11.289 - 0.77695x = 12.37 0.1285x= 1.08097 x= 8.409g 8.409g Percent BaO=� 100 = 57.87% = 57.9% Percent BaCO 42.1% 214.53g 3

23. 2 NH3 + 3 CuO  N2 + 3 Cu + 3 H2O

1mol NH3 1mol N2 28.013g Mass nitrogen gas formed: from NH3 =18.1g NH 3 创 � 14.88g 17.030g 2mol NH3 1mol N 2 1mol CuO1mol N 28.013g from CuO=90.4g CuO创 2 � 10.61g 79.545g 3mol CuO 1mol N2

CuO is limiting, NH3 is in excess and 10.6g of nitrogen is formed

1mol CuO2mol NH3 17.030g Mass NH3 used=90.4g CuO创 � 12.90g 79.545g 3mol CuO 1mol NH3

Mass NH3 remaining = 18.1 g - 12.9 g = 5.2 g Chem 121 Problem set III Solutions - 6

24. 2 XeF2 + 2 H2O ® 2 Xe + 4 HF + O2 1.00g 50.0g mass of HF formed

1mol XeF2 4mol HF 20.01g from XeF2 = 1.00g创 � 0.23639g 169.3g 2mol XeF2 1mol HF

1mol H2 O 4mol HF 20.01g from H2 O= 50.0g创 � 111.0g 18.02g 2mol H2 O 1mol HF

thus XeF2 is limiting and H2O is in excess mass HF formed is 0.236g

1mol XeF2 2mol H 2 O 18.02g mass of H2 O used=1.00g创 � 0.1064g 169.3g 2mol XeF2 1mol H 2 O

mass of H2O remaining = 50.0g - 0.1064g = 49.9g

25. 4 NH3 (g) + 5 O2(g)  4 NO(g) + 6 H2O(l) 2.00g 4.50g

mass of NO formed

1mol NH3 4mol NO 30.01g from NH3 = 2.00g创 � 3.524g 17.03g 4mol NH3 1mol NO

26. This problem can be solved in a couiple of ways, but the easiest is if it is treated as a empirical formula.

Percentage of MgSO4 in hydrate = 100  51.2 = 48.8 So take 100g of hydrate molecule Mass Moles Divide by Closest whole number ratio (g) smallest

MgSO4 48.8 0.40545 1 1

H2O 51.2 2.84128 7.00 7

and so there are 7 molecules of water for every molecule of MgSO4 in the hydrate.

limiting reagent is AgNO3 , KCl is in excess and 4.2g AgCl is formed. 1mol AgNO 1mol KCl 74.55g mass KCl used= 5.0g创 3 � 2.194g 169.91g 1mol AgNO3 1mol KCl mass KCl remaining = 5.0g - 2.194g = 2.8g

1mol O2 4mol NO 30.01g from O2 = 4.50g创 � 3.376g 32.00g 5mol O2 1mol NO

thus O2 is limiting and NH3 is in excess

mass NO formed is 3.38g

1mol O2 4mol NH3 17.03g Mass NH3 used= 4.50g创 � 1.915g 32.00g 5 mol O2 1mol NH 3

mass of NH3 remaining = 2.00g - 1.92g = 0.08g

27. Percentage Yield = 1.21g * 100 / 1.25 g = 96.8%

st 28. - 1 write the equation: N2 + H2  NH3

- then balance it: N2 + 3H2  2NH3

- then determine limiting reactant:

 with this easy an equation, can see that for 4 mol N2 would need 3 x 4 = 12 mol H2   H2 is limiting Chem 121 Problem set III Solutions - 7

- then calculate the theoretical yield on the basis of 6.0 mol H2:

骣2 mol NH3 mol NH3=( 6.0 mol H 2)琪 = 4.0 mol NH 3 桫3 mol H2 骣 actual yield骣 1.6 � then calculate %yield=琪 ( 100%) = 琪 (100%) 40% 桫theoretical yield桫 4.0

29. a) C6H6 + Br2  C6H5Br + HBr 30.0g 65.0g

1mol C6 H 6 1mol C 6 H 5 Br 157.01g mass of C6H5Br formed from C6 H 6 = 30.0g创 � 60.30g 78.11g 1mol C6 H 6 1mol C 6 H 5 Br

1mol Br2 1mol C6 H 5 Br 157.01g from Br2 = 65.0g创 � 63.86g 159.81g 1mol Br2 1mol C 6 H 5 Br

thus C6H6 is limiting and Br2 is in excess mass C6H5Br formed is 60.3g

1mol C6 H 6 1mol Br2 159.81g Mass Br2 used= 30.0g创 � 61.38g 78.11g 1mol C6 H 6 1mol Br 2

mass of Br2 remaining = 65.0g - 61.4g = 3.6g 56.7g b) Percentage Yield=� 100 94.0% 60.3g

30. - 1st we write out the equation:

H2O + KO2  KOH + O2 - 2nd we balance it:

2H2O + 4KO2  4KOH + 3O2

- then we determine which reactant is limiting: i) assume H2O is limiting 骣 骣 1mol H2 O 4 mol KOH骣 56.11g KOH mol KOH=( 25.0 g H2 O)琪 琪 琪 = 156 g KOH 桫18.02 g H2 O 桫 2 mol H 2 O桫 1mol KOH

ii) assume KO2 is limiting: 骣 骣 1mol KO2 4 mol KOH骣 56.11g KOH mol KOH=( 25.0 g KO2 )琪 琪 琪 = 19.7 g KOH 桫71.10 g KO2 桫 4 mol KO 2 桫 1mol KOH

 so KO2 is limiting since it forms the least amount of product  and the theoretical yield = 19.7 g KOH

骣 15.3 g KOH actual - 3rd calculate the % yield: %yield=琪 (100%) = 77.7% 桫19.7 g KOH theoretical

31. - we’re told how much iron is formed in the 2nd reaction  we need to calculate the amount of CO required to get this amount of Fe, using the molar relationships from the 2nd equation  then we can calculate how much carbon is required to generate that much CO from the first equation 骣1mol Fe 骣 3 mol CO mol CO=( 750.0 g Fe)琪 琪 = 20.14 mol CO 桫55.85 g Fe 桫 2 mol Fe 骣2 mol C 骣 12.01g C mass C=( 20.14 mol CO)琪 琪 = 241.9 g C 桫2 mol CO 桫 1mol C Chem 121 Problem set III Solutions - 8

骣12.0 g NaCl 32. mass NaCl=( 125.0 g soln)琪 = 15.0 g NaCl 桫100 g soln

骣100 g soln 骣 1mL soln 33. mL soln=( 75.0 g NaCl)琪 琪 = 451mL soln 桫15.0 g NaCl 桫 1.108 g soln

骣 1mol KNO3 34. mol KNO3=( 3.765 g KNO 3)琪 = 0.03724 mol KNO 3 桫101.10 g KNO3

骣0.03724 mol KNO3 骣1000 mL soln [KNO3 ]=琪 琪 = 0.1241M KNO 3 桫300.0 mL soln 桫 1L soln

骣1mol NaNO3 35. mol NaNO3=( 20.00 g NaNO 3)琪 = 0.2353 mol NaNO 3 桫85.00 g NaNO3 骣1mL soln 骣 1L vol soln=( 100.0 g soln)琪 琪 = 0.08749 L soln 桫1.143 g soln 桫 1000 mL 0.2353 mol NaNO [NaNO ]=3 = 2.689 M NaNO 30.08749 L soln 3

M1 V 1 25.00 mL x 0.3447 M M2 = = = 0.00862 M V2 1000.0 mL 36. M1V1 = M2V2

骣2.00 mol K2 SO 4 37. mol K2 SO 4=( 0.0145 L)琪 = 0.0290 mol K 2 SO 4 桫 1L 0.0290 mol K SO [K SO ]=2 4 = 0.0967 M K SO 2 40.300 L soln 2 4

38. - we know M1 and V1 and M2  we solve the dilution eqn for V2 and subtract the initial volume of HCl soln from final volume to determine the amount of water to add:

M1 V 1 (12.0 M)( 5.00 mL) V2 = = = 100 mL soln M2 ( 0.600 M) water added = 100 mL soln required – 5.00 mL 12.0 M HCl = 95 mL water added

st 39. - 1 calculate the # of moles of AgNO3 present  then find the # of moles of K2CrO4 to react with it  finally, use molarity to find the volume of interest

骣0.420 mol K2 CrO 4 骣2 mol AgNO3骣 169.98 g AgNO 3 mass AgNO3=( 75.00 mL K 2 CrO 4 soln)琪 琪 琪 = 10.71g AgNO 3 桫 1000 mL soln桫 1mol K2 CrO 4桫 1mol AgNO 3 moles MnO - 1 L 40. moles MnO- reacted = 0.0200 4 x 22.4 mL MnO - x = 4.48 x 10-4 mole 4L 4 1000 mL 2+ 2+- -4 - 5 mole Fe -3 moles of Fe reacted with MnO4 = 4.48 x 10 mole MnO 4 x - = 2.24 x 10 mole 1 mole MnO4 2.24 x 10-3 mole 1000 mL [Fe2+ ] = x = 0.056 M 40.0 mL Fe2+ 1 L

41.  determine the # mol KMnO4 , then mol Na2C2O4 required for the reaction Chem 121 Problem set III Solutions - 9

 then use definition of percentage 骣 骣 0.08395 mol KMnO4 5 mol Na 2 C 2 O 4 mass Na2 C 2 O 4= ( 18.74 mL KMnO 4 soln)琪 琪 桫1000 mL KMnO4 soln 桫 2 mol KMnO 4

骣134.00 g Na2 C 2 O 4 x琪 = 0.5270 g Na2 C 2 O 4 桫 1mol Na2 C 2 O 4

骣0.5270 g Na2 C 2 O 4 %Na2 C 2 O 4=琪 ( 100%) = 13.95 %Na 2 C 2 O 4 桫 3.778 g sample

42. H2SO4 + 2 NaOH  Na2SO4 + 2 H2O 1L mol 1mol H SO moles H SO =32.74mL NaOH创 2.15� 2 4 0.035195 2 4 1000mL L 2mol NaOH 0.035195mol H SO 1000mL mol [H SO ]=2 4 � = 3.5195 3.52M 2 4 10.00mL 1L L 43. - convert mass KHP to moles KHP  which converts to mole NaOH and  M NaOH soln 骣1mol KHP 骣 1mol NaOH mol NaOH=( 0.7284 g KHP)琪 琪 = 3.567 x 10-3 mol NaOH 桫204.2 g KHP 桫 1mol KHP 3.567 x 10-3 mol NaOH [NaOH]= = 0.1032 M NaOH 0.03458 L NaOH soln

44. Step 1. Balance the equations 2Cu2O  4Cu + O2

2CuO  2Cu + O2

Step 2. Define the variables for CuO and Cu2O. Let the amount of Cu2O be X g and so the amount of CuO is (1.000 - X) g.

Step 3: Mass of Cu formed from:

1 mole Cu2 O 4 mole Cu 63.55 g Cu Cu2 O: X g Cu 2 O x x x = 0.88819X 143.1 g Cu2 O 2 mole Cu 2 O 1 mole Cu 1 mole CuO 2 mole Cu 63.55 g Cu CuO: (1.000-X) g CuO x x x = 0.79887(1.000-X) 79.55 g CuO 2 mole CuO 1 mole Cu So the total mass of Cu formed is 0.88819X g + 0.79887(1.000-X) g = 0.8390 g

Step 4: Solve for X X = 0.4493 g = mass of Cu2O 0.4493 g Therefore percent Cu O in mixture is x 100 = 53.6 % 2 0.8390 g

45. Let mass of Mg = x , so mass of Zn = (1.000 - x) g

Mg + 1/2 O2 ® MgO

Zn + 1/2 O2 ® ZnO 1mol Mg 1mol MgO 40.31g mass of MgO from Mg=x g创 � 1.6582x g 24.31g 1mol Mg 1mol MgO 1mol Zn 1mol ZnO 81.38g mass of ZnO from Zn=(1.000-x) g创 � 1.2447(1.000x) g 65.38g 1mol Zn 1mol ZnO but mass of oxide mixture = 1.409 g = 1.6582x+1.2447(1.000-x) 1.2447- 1.2447x + 1.6582x = 1.409 0.41346x= 0.1643 Chem 121 Problem set III Solutions - 10

x= 0.3973g = 0.397g Percentage of Zn in the mixture is 60.3% , and percent Mg is 39.7%

46. - start with an unbalanced equation to get an idea of what’s going on:

KCl + MgCl2 + H2SO4  HCl - we can write a balanced equation for the titration:

HCl + NaOH  H2O + NaCl  we know that mol NaOH = mol HCl = mol Cl 骣0.1054 mol NaOH mol Cl=( 75.82 mL NaOH)琪 = 7.991 x 10-3 mol NaOH = mol Cl 桫1000 mL NaOH soln 骣35.45 g Cl mass Cl=( 7.991 x 10-3 mol Cl)琪 = 0.2833 g Cl 桫1mol Cl

- we also know that: g KCl + g MgCl2 = 0.502 g  so let x = g KCl, then g MgCl2 must be (0.502 –x)  then we can calculate the mol of Cl that each compound contributes

骣1mol KCl 骣 1mol Cl 骣 35.45 g Cl mass Cl from KCl=( x g KCl)琪 琪 琪 = 0.4762x g Cl 桫74.55 g KCl 桫 1mol KCl 桫 1mol Cl 骣 骣 1mol MgCl2 2 mol Cl骣 35.45 g Cl mass Cl from MgCl2=( x g MgCl 2 )琪 琪 琪 = 0.7447(0.502 - x) g Cl 桫95.21g MgCl2 桫 1mol MgCl 2 桫 1mol Cl

- the 2 compounds added together donated all the Cl in the HCl formed:  0.4762x + 0.7447(0.502 – x) = 0.2833 g Cl

0.4762x + 0.3738 - 0.7447x =0.2833

0.2685x = .09050 and x = 0.337 = g KCl and  g MgCl2 = 0.502 – 0.337 = 0.165 g MgCl2

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