Download + H 2 O(g)

1) REACTIONs:
a)
Al4C3(s) + H2O (l)=> Al(OH)3 (s) + CH4 (g)
Balance the reaction. Describe the chemical process represented by this reaction. Write the name of each
single reactant and product.
Solution :
First, the oxidation number of each atom have to be
calculated and check if any of them changed. None of the
oxidation numbers changes. It is not a redox reaction. The
aluminium carbide reacts with water to form the aluminium
hydroxide and methane. Then we can balance by an
intuitive method. It is an heterogeneous phase reaction, as
the carbide is a solid, the aluminium hydroxide is not
soluble in water and methane is gaseous then it evolves as
formed.
Al4C3(s) + 12 H2O (l)=> 4Al(OH)3 (s) + 3CH4 (g)
Al4C3 = aluminium carbide
H2O = hydrogen oxide
Al(OH)3 = aluminium hydroxide
CH4 = methane.
b)
CaO(s) + NH4Cl(s) → NH3(g) + H2O(g) + CaCl2 (s)
Balance the reaction, describe which type of reaction is.
Solution :
First, the oxidation number of each atom have to be
calculated and check if any of them changed. None of the
oxidation numbers changes. CaO is a basic oxide as well as
NH4+ is the protonated ammonia, then it can deliver H+ as
an acid, then this reaction might be an acid base reaction.
It occurs in heterogeneous phases, mostly in the solid phase
with the formation of two gases, NH3 and water. We can
assume that high temperature can be reached as the water
is a gas and not a liquid as normally is. We can suppose that
the reaction is exothermic. We can observe that the
balancing is through an intuitive method.
CaO(s) + 2NH4Cl(s) → 2NH3(g) + H2O(g) + CaCl2 (s)
c)
Na2SO4 (aq) + BaCl2 (aq) => BaSO4(s) + NaCl(aq)
Balance the reaction. Describe to which class of compounds (oxide, hydride, acid, base, salt) the reactants
and the products belong. Write the ionic and net ionic equation.
Solution :
First, the oxidation number of each atom have to be
calculated and check if any of them changed. None of the
oxidation numbers changes. It is a double exchange
reaction where the driving force is the formation of an
insoluble product as the BaSO4. Checking the table of the
solubility (if it is available) we observe that the sulfate salts
of earth alkaline metals are always insoluble. The balancing
is based on the intuitive method.
Na2SO4 (aq) + BaCl2 (aq) => BaSO4(s) + 2NaCl(aq)
Describe to which class of compounds (oxide, hydride, acid, base, salt) the reactants and the products belong
Na2SO4 = sodium sulfate
BaCl2 = barium chloride
BaSO4= barium sulfate
NaCl= sodium chloride
(salt)
(salt)
(salt)
(salt)
d) Depict with a x, which one of the following reactions occurs as they are written or they do not occur:
Zn + 2HCl => ZnCl2 + H2

2Na + 2H2O = 2NaOH + H2

2Au + 6HNO3 => 2Au(NO3)3 + 3H2

Ni + 2H2O => Ni(OH)2 + H2

Be + 2H2O => Be(OH)2 + H2

Solution :
First, these reaction are single exchange reactions and they
are redox processes because the formation of molecular
hydrogen in observed for all of them. By checking the
oxidation numbers it is clear that the metals are oxidized
and the hydrogen reduced.
To answer to these questions we need to recall the
concepts of hydrogen displacement by the metals.
Now we can answer:
Zn + 2HCl => ZnCl2 + H2
2Na + 2H2O = 2NaOH + H2
2Au + 6HNO3 => 2Au(NO3)3 + 3H2
Ni + 2H2O => Ni(OH)2 + H2
Mg + 2H2O =>Mg(OH)2 + H2
x
x


x (if water is heated to steam)
2) Analysis of a gaseous chlorofluorocarbon, CClxFy, shows that it contains 9.93 % of C and 58.64 % of Cl. In another
experiment you find that 0.107 g of the compound fills a 458 mL flask at 25°C with a pressure of 21.3 mmHg .
what is the molecular formula of the compound?
Solution :
First, according the definite proportion law we have to
calculate the empirical formula.
100- 58.64-9.93 = 31.43% of fluorine
9.93 /12.01 = 0.8275
58.64/35.45 = 1.6542
31.43/18.99 = 1.6551
1.6542/0.8275 = 2
Empyrical formula = CCl2F2
0.107 g of the compound fills a 458 mL flask at 25°C with a pressure of 21.3 mmHg
21.3 mmHg /760 mmHg /atm = 0.028 atm
25 + 273 = 298 K
458 mL = 0.458 L
𝑃𝑣 = 𝑛𝑅𝑇 =
𝑀𝑀 =
𝑀𝑀 =
𝑔
𝑅𝑇
𝑀𝑀
𝑔𝑅𝑇
𝑃𝑉
0.0823 𝐿 𝑎𝑡𝑚
298𝐾
𝑚𝑜𝑙 𝐾
= 204.84 𝑔/𝑚𝑜𝑙
0.028 𝑎𝑡𝑚 0.458𝐿
0.107 𝑔
The molar mass of the empirical formula is :
12.011 + 35.45 x 2 + 18.99 x 2= 204.9
The empirical formula represents the molecular formula:
CCl2F2
3) According to the following reaction:
CaO(s) + NH4Cl(s) => NH3 (g)+ H2O + CaCl2(s)
If 112g of CaO and 224 g of NH4Cl are mixed we obtain 16.3 g of gaseous ammonia. What is the yield of the
reaction? The gas was collected in a 2L flask filled with water. How many moles of ammonia will be in 200 mL
of this solution?
Solution :
First, according to the reaction equation we balance the reaction.
The oxidation numbers do not change. It is an acid (NH4+) and
base (CaO) reaction. We balance by an intuitive method.
CaO(s) + 2NH4Cl(s) => 2NH3 (g)+ H2O + CaCl2(s)
If 112g of CaO and 224 g of NH4Cl are mixed we obtain 16.3 g of gaseous ammonia. What is the yield of the
reaction?
Being CaO and NH4Cl reactants we have to check if they are
in the right molar ratio.
112 g /MM CaO = 112 g / 56.08 g/mol = 1.997 mol
224 g/MM NH4Cl = 224 / 53.49 g/mol = 4.188 mol
From the reaction the mole ratio between CaO and NH4Cl is 1
: 2 and we have almost 1 : 4. It is clear that CaO is the limiting
reactant.
Then the mole ratio between CaO and NH3 in the reaction is
1:2 then when it reacts 1.997 mol of CaO 1.997 mol x 2 are
the mole of NH3.
1.997 mol x 2 = 3.994 mol
3.994 mol x MM NH3 =
= 3.994 mol x 17.03 g/mol = 68.02 g theoretical mass of NH3.
To calculate the yield the proportion can be set up:
16.3 : 68.02 = x : 100
X = 23.96 % yield of the reaction
4) A mixture of CuCl2 and CuCl2 . 2H2O has a mass of 1.565 g. After heating to drive off all the water, the mass is
only 1.391 g. What is the mass percent of CuCl2 . 2H2O in the mixture?
Solution :
First, we have to consider the relationship between the two
salts; one contains water and the other is dry and they are
mixed in a proportrion which is unknown.
1.565 – 1.391 = 0.174 g of H2O
1.391 g / MM CuCl2 = 1.391 g/ 134.45 g/mol = 0,01034 moles of
dry CuCl2
0.174 g / MM H2O = 0.174 g / 18.02 g/mol = 0.09666 moles of
water
If 0.09666 are the moles of H2O, the mole ratio in the formula
CuCl2 . 2 H2O between CuCl2 and H2O is 1 : 2, respectively.
Hence, 0.09666/ 2 = 0.04833 are the moles of CuCl2 in the
formula CuCl2 . 2 H2O
0.04833 moles x MM CuCl2 = 0.04833 moles x 134.45 g/mol =
0.6498 g Mass of CuCl2 due to CuCl2 x 2 H2O
Mass of CuCl2 x 2 H2O = 0.6498 g (CuCl2) + 0.174 g (H2O)=
0.8238 g
What is the mass percent of CuCl2 . 2H2O in the1.565 g of the mixture?
0.8238 : 1.565 = y = 100
Y = 52.64%