Chem 112, Answers to Problem Set II
Chem 121 Problem set III Solutions - 10
Problem Set III Stoichiometry - Solutions
1.
2.
3. molecular mass of ethane = 2(12.011) + 6(1.008) = 30.07 g
4. molecular mass of aniline = 6(12.011) + 7(1.008) + 14.01= 93.14 g/mol
5. Let the molar mass of X be x.
x = 31.1 g/mol and the element is P.
6. Let atomic mass of M be x
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
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
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
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
® 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
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
percent C =
moles H2O ® moles H in H2O ® moles H in compound ® mass H in compound ® percent H
percent H =
Percentage of O in unknown =
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 numberC / 40.0 / /
H / 6.72 / /
O / 53.3 / /
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.
The ratio of molecular mass to empirical mass is
So the molecular formula is (CH2O)6 or C6H12O6.
11. Take a 100.00 g sample
Atom / Mass (g) / Moles / Divide by smallest / Best ratioNa / 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
12.
Element / Relative mass / Relative number of moles (atoms) / Divide by the smallest numberLa / 0.27004 / /
O / 0.046661 / /
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
\ the % of Ti in the product is:
and all the rest is S ® 100 – 42.86 = 57.14 %S
b. for the empirical formula, require moles
® sulphur was in excess, \ use the % S in the product to get the amount of sulphur that must have reacted:
or could have got g S from 20.58 g – 8.820 g = 11.76 g S
and
Ti0.184/0.184 S0.367/0.184 ® Ti1S1.99 ~ TiS2
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
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 moles (atoms) / Divide by the smallest number / Whole numberratio
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
- now balance it: C4H10 + 13/2O2 ® 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
19.
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
21. 2 C57H110O6 + 163 O2 ® 114 CO2 + 110 H2O
22. Let mass BaO2 = (x) g , thus mass BaCO3 = (14.53 - x)g
Mass BaO formed:
23. 2 NH3 + 3 CuO ® N2 + 3 Cu + 3 H2O
Mass nitrogen gas formed:
CuO is limiting, NH3 is in excess and 10.6g of nitrogen is formed
Mass NH3 remaining = 18.1 g - 12.9 g = 5.2 g
24. 2 XeF2 + 2 H2O ® 2 Xe + 4 HF + O2
1.00g 50.0g
mass of HF formed
thus XeF2 is limiting and H2O is in excess
mass HF formed is 0.236g
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
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 (g) / Moles / Divide by smallest / Closest whole number ratioMgSO4 / 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.
mass KCl remaining = 5.0g - 2.194g = 2.8g
thus O2 is limiting and NH3 is in excess
mass NO formed is 3.38g
mass of NH3 remaining = 2.00g - 1.92g = 0.08g
27. Percentage Yield = 1.21g * 100 / 1.25 g = 96.8%
28. - 1st 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
- then calculate the theoretical yield on the basis of 6.0 mol H2:
29. a) C6H6 + Br2 ® C6H5Br + HBr
30.0g 65.0g
mass of C6H5Br formed
thus C6H6 is limiting and Br2 is in excess
mass C6H5Br formed is 60.3g
mass of Br2 remaining = 65.0g - 61.4g = 3.6g
b)
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
ii) assume KO2 is limiting:
® so KO2 is limiting since it forms the least amount of product
® and the theoretical yield = 19.7 g KOH
- 3rd calculate the % yield:
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
36. M1V1 = M2V2
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:
water added = 100 mL soln required – 5.00 mL 12.0 M HCl = 95 mL water added
39. - 1st 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
40.
41. ® determine the # mol KMnO4 , then mol Na2C2O4 required for the reaction
® then use definition of percentage
42. H2SO4 + 2 NaOH ® Na2SO4 + 2 H2O
43. - convert mass KHP to moles KHP ® which converts to mole NaOH and \ M 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:
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
Therefore percent Cu2O in mixture is
45. Let mass of Mg = x , so mass of Zn = (1.000 - x) g
Mg + 1/2 O2 ® MgO
Zn + 1/2 O2 ® ZnO
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
- 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
- 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