TUTORIAL #2 – Mass Relationship in Chemical Reactions (part I):
1.
Vanillin is the compound responsible for the vanilla flavor and its structure is shown below.
Calculate the percentage of carbon in vanillin
Answer:
%𝐶 =
2.
8 × 12.01
× 100% = 63.15 %
152.1422
The molecular formula of allicin, the compound responsible for the characteristic smell of
garlic, is C6H10OS2.
a) What is the molar mass of allicin?
b) How many moles of allicin are present in a sample of 5.00 mg?
c) How many molecules are there in 5.00 mg of allicin?
d) How many S atoms are present in 5.00 mg of allicin?
Answer:
a)
(6 x 12.01 g/mol) + (10 x 1.008 g/mol) + 15.9994 g/mol + (2 x 32.066 g/mol) =162.2714 g/mol
b) 5.00 𝑚𝑔 ×
1𝑔
1000 𝑚𝑔
c) 2.97 × 10−5 𝑚𝑜𝑙 ×
×
1 𝑚𝑜𝑙 𝑎𝑙𝑙𝑖𝑐𝑖𝑛
162.2714 𝑔
6.022 × 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
1 𝑚𝑜𝑙
d) 1.79 × 1019 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 ×
3.
= 3.08 × 10−5 𝑚𝑜𝑙
2 𝑆 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
= 1.86 × 1019 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
= 3.71 × 1019 𝑆 𝑎𝑡𝑜𝑚𝑠
Determine the empirical formula of the compounds with the following composition by
mass:
a) 10.4% C, 27.8% S and 61.7% Cl;
b) 32.79% Na, 113.02% Al and 54.19% F;
Answer: (a)CSCl2 (b) Na3AlF6
IUPAC name: Carbonothioyl dichloride
Other names: Thiophosgene; Thiocarbonyl
chloride, Carbonothioic dichloride
Trisodium Hexafluoroaluminate
Aluminum trisodium hexafluoride
Sodium aluminum hexafluoride
Sodium fluoaluminate(3-)
Sodium hexafluoroaluminate (Na3AlF6)
Trisodium aluminum hexafluoride
Sodium hexafluoroaluminate
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4.
Balance the following equations:
a) Mg + HNO3  H2 + Mg(NO3)2
b) CaC2 + H2O  Ca(OH)2 + C2H2
c) S + O2  SO3
d) UO2 + HF  UF4 + H2O
Answer:
a) Mg + 2 HNO3  H2 + Mg(NO3)2
b) CaC2 + 2 H2O  Ca(OH)2 + C2H2
c) 2 S + 3 O2  2 SO3
d) UO2 + 4 HF  UF4 + 2 H2O
5.
In the preparation of iron from hematite, Fe2O3 reacts with carbon:
Fe2O3 + C  Fe + CO2 (unbalanced)
a) Balance the equation.
b) How many grams of carbon are needed to react with 525 g of hematite?
Answer:
a) 2 Fe2O3 + 3 C  4 Fe + 3 CO2
b) 525 𝑔 𝐹𝑒2 𝑂3 ×
6.
1 𝑚𝑜𝑙 𝐹𝑒2 𝑂3
159.69 𝑔 𝐹𝑒2 𝑂3
×
3 𝑚𝑜𝑙 𝐶
2 𝑚𝑜𝑙 𝐹𝑒2 𝑂3
×
12.01 𝑔 𝐶
1 𝑚𝑜𝑙 𝐶
= 59.23 𝑔 𝐶
Magnesium metal burns in oxygen to form magnesium oxide.
a) Write a balance equation for the reaction.
b) How many grams of magnesium oxide will be produced from 25.0 g magnesium metal?
Answer:
a) 2 Mg + O2  2 MgO
b) 25.0 𝑔 𝑀𝑔 ×
7.
1 𝑚𝑜𝑙 𝑀𝑔
24.305 𝑔 𝑀𝑔
×
2 𝑚𝑜𝑙 𝑀𝑔𝑂
2 𝑚𝑜𝑙 𝑀𝑔
×
40.3044 𝑔 𝑀𝑔𝑂
1 𝑚𝑜𝑙 𝑀𝑔𝑂
= 41.46 𝑔 𝑀𝑔𝑂
Hydrogen and chlorine react to yield hydrogen chloride: H2 + Cl2  2 HCl
How many grams of HCl are formed from the reaction of 2.56 g of H2 with 8.94 g of Cl2?
Which is the limiting reactant?
Answer:
1 𝑚𝑜𝑙 𝐻
If all H2 is used up: 2.56 𝑔 𝐻2 × 2.016 𝑔 𝐻2 ×
2
If all Cl2 is used up: 8.94 𝑔 𝐶𝑙2 ×
1 𝑚𝑜𝑙 𝐶𝑙2
70.9 𝑔 𝐶𝑙2
×
2 𝑚𝑜𝑙 𝐻𝐶𝑙
1 𝑚𝑜𝑙 𝐻2
×
36.458 𝑔 𝐻𝐶𝑙
1 𝑚𝑜𝑙 𝐻𝐶𝑙
= 92.59 𝑔 𝐻𝐶𝑙
2 𝑚𝑜𝑙 𝐻𝐶𝑙
1 𝑚𝑜𝑙 𝐶𝑙2
×
36.458 𝑔 𝐻𝐶𝑙
1 𝑚𝑜𝑙 𝐻𝐶𝑙
= 9.19 𝑔 𝐻𝐶𝑙
Therefore, limiting reactant is Cl2 and 9.19 g HCl will be formed.
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8.
You are given 1.39 mol of H2 and 3.44 mol of N2. How many grams of ammonia can you
make? How many grams of excess reactant will be left over?
3 H2 + N2  2 NH3
Answer:
If all H2 is used up: 1.39 𝑚𝑜𝑙 𝐻2 ×
2 𝑚𝑜𝑙 𝑁𝐻3
3 𝑚𝑜𝑙 𝐻2
= 0.927 𝑚𝑜𝑙 𝑁𝐻3
If all N2 is used up: 3.44 𝑚𝑜𝑙 𝑁2 ×
2 𝑚𝑜𝑙 𝑁𝐻3
1 𝑚𝑜𝑙 𝑁2
= 6.88 𝑚𝑜𝑙 𝑁𝐻3
Therefore, limiting reactant is H2
and grams of NH3 made = 0.927 𝑚𝑜𝑙 𝑁𝐻3 ×
17.031 𝑔𝑁𝐻3
1 𝑚𝑜𝑁𝐻3 𝑙
= 15.79 𝑔 𝑁𝐻3
mass of leftover of N2 = [3.44 𝑚𝑜𝑙 𝑁2 − (1.39 𝑚𝑜𝑙 𝐻2 ×
1 𝑚𝑜𝑙 𝑁2
)]
3 𝑚𝑜𝑙 𝐻2
×
28.014 𝑔 𝑁2
1 𝑚𝑜𝑙 𝑁2
= 83.39 𝑔 𝑁2
TUTORIAL #2 – Mass Relationship in Chemical Reactions (part II):
9.
Predict whether a precipitation reaction will occur when aqueous solutions of the following
substances are mixed. If precipitation reaction occurs, then write the molecular equation,
ionic equation and net ionic equation for it.
a) NaOH + HClO4
b) FeCl2 + KOH
c) (NH4)2SO4 + NiCl2
d) CH3COONa + HCl
Answer:
a) NaOH + HClO4  NaClO4 + H2O
(no precipitation)
b) FeCl2 + 2 KOH  Fe(OH)2 + 2 KCl
(Fe(OH)2 is insoluble)
Molecular equation: FeCl2 + 2 KOH  Fe(OH)2 + 2 KCl
Ionic equation: Fe2+ + 2 Cl– + 2 K+ + 2 OH–  Fe(OH)2 + 2 K+ + 2 Cl–
Net ionic equation: Fe2+ + 2 OH–  Fe(OH)2
c) (NH4)2SO4 + NiCl2  NH4Cl + NiSO4
(no precipitation)
d) CH3COONa + HCl  CH3COOH + NaCl
10.
(no precipitation)
Which element is oxidized and which is reduced in the following reactions?
a) Ca + Sn2+  Ca2+ + Sn
b) ICl + H2O  HCl + HOI
c) Si + 2 Cl2  SiCl4
d) Cl2 + 2 NaBr  Br2 + 2 NaCl
3
Answer:
a)
b)
c)
d)
11.
Ca is oxidized & Sn2+ is reduced
It is not a redox reaction, no substance is oxidized or reduced
Si is oxidized & Cl2 is reduced
Br– is oxidized & Cl2 is reduced
How many moles of solute are present in 35.0 mL of 1.200 M HNO3?
Answer: 35.0 𝑚𝐿 ×
12.
1𝐿
1000 𝑚𝐿
×
1.200 𝑚𝑜𝑙
1𝐿
The sterile saline solution used to rinsed contact lenses can be made by dissolving 400 mg of
NaCl in sterile water and diluting to 100 mL. What is the molarity of the solution?
1𝑔
Answer: (400 𝑚𝑔 𝑁𝑎𝐶𝑙 × 1000 𝑚𝑔 ×
13.
1 𝑚𝑜𝑙 𝑁𝑎𝐶𝑙
)
58.45 𝑔 𝑁𝑎𝐶𝑙
×
1
0.100 𝐿
= 0.068 𝑀
A bottle of 12.0 M HCl has only 35.7 mL left in it. What will the concentration be if the
solution is diluted to 250.0 mL?
Answer:
12.0 𝑀 × 35.7 𝑚𝐿 = 𝑀2 × 250.0 𝑚𝐿
∴ 𝑀2 =
14.
= 0.042 𝑚𝑜𝑙
12.0 𝑀 × 35.7 𝑚𝐿
= 1.7 𝑀
250.0 𝑚𝐿
A flask containing 450 mL of 0.500 M HBr was accidentally knocked to the floor. How many
grams of K2CO3 would you need to put on the spill to neutralize the acid according to the
following equation?
2 HBr + K2CO3  2 KBr + CO2 + H2O
Answer:
15.
450 ×0.500
1000
𝑚𝑜𝑙 𝐻𝐵𝑟 ×
1 𝑚𝑜𝑙 K2 𝐶𝑂3
2 𝑚𝑜𝑙 𝐻𝐵𝑟
×
138.206 𝑔 K2 𝐶𝑂3
1 𝑚𝑜𝑙 K2 𝐶𝑂3
= 15.55 𝑔 K 2 𝐶𝑂3
An unidentified metal M reacts with an unidentified halogen X to form a compound MX2.
When heated, the compound decomposes by the reaction:
2 MX2 (s)  2 MX (s) + X2 (g)
When 1.12 g of MX2 is heated, 0.720 g of MX is obtained, along with 56.0 mL of X2 gas.
Under the condition used, 1.00 mol of the gas has a volume of 22.41 L.
a) What is the atomic mass and identity of the halogen X?
b) What is the atomic mass and identity of the metal M?
Answer: (a) 80 amu, Br (b) 64 amu, Cu
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