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CHEM1001 Answers to Problem Sheet 5
1
...
9 g and consists of 6
...
The mass of
12 × 1023 atoms is therefore:
mass =

𝟏𝟐×𝟏𝟎𝟐𝟑
𝟔
...
9 g mol-1 = 220 g

As the number of atoms is only known to 2 significant figures, this mass should be
rounded to 220 g
...
003 g so 6 moles has a mass of 6 × 4
...
018 g
...
00 has two trailing zeros
indicating that it has been measured this accurately)
...
0 g
...
0 g) < copper (100 g) < silver (220 g)
2
...
01 (C)) + (4 × 1
...
052 g mol-1
The number of moles in 112 g of C2H4 is therefore:
number of moles of C2H4 =

𝐦𝐚𝐬𝐬
𝐦𝐨𝐥𝐚𝐫  𝐦𝐚𝐬𝐬

=

𝟏𝟏𝟐  𝐠
𝟐𝟖
...
99 mol

The mass of 112 g is known to 3 significant figures and so the answer is also given to this
level of accuracy
...
Each mole of C2H4
is made up of 2 moles of C and 4 moles of H:
number of moles of C = 2 × 3
...
98 mol
Again, the figure has been rounded to reflect the three significant figures given for the
mass of ethylene
...


The molar mass of C4H10 is ((4 × 12
...
008 (H))) g mol-1 = 58
...
0 g of C4H10 is therefore:
number of moles of C4H10 =

𝐦𝐚𝐬𝐬
𝐦𝐨𝐥𝐚𝐫  𝐦𝐚𝐬𝐬

The equation for the combustion is:
C4H10(g) +

13
O2 à 4CO2(g) + 5H2O(l)
2

=

𝟏𝟎
...
𝟏𝟐  𝐠  𝐦𝐨𝐥!𝟏

= 0
...

2
(i)

The number of moles of CO2 produced is therefore 4 × 0
...
688 mol
...
01 (C) + 2 × 16
...
01 g mol-1
...
688 mol of CO2 is:
mass of CO2 = number of moles × molar mass = 0
...
01 g mol-1 = 30
...


(ii)

The number of moles of H2O produced is 5 × 0
...
860 mol
...
008 (H) + 16
...
016 g mol-1
...
860 mol of H2O is:
mass of H2O = number of moles × molar mass = 0
...
016 g mol-1 = 15
...


(iii)

13
× 0
...
118 mol
...
00) g mol-1 = 32
...
The mass of 1
...
118 mol × 32
...
8 g
...


Sodium carbonate is Na2CO3 and has a formula mass of
formula mass = (2×22
...
01 (C) + 3×16
...
99 g mol-1
The number of moles in 53
...
𝟎  𝐠
𝟏𝟎𝟓
...
500 mol

The reaction is:
2NaOH + CO2 à Na2CO3 + H2O
Two moles of NaOH are required for every mole of Na2CO3 that is prepared
...
500
moles of Na2CO3 therefore requires 1
...

Each mole of NaOH has a mass of 22
...
00 (O) + 1
...
998 g mol-1
...
00 mol of NaOH is:
mass of NaOH = number of moles × molar mass
= (39
...
00 g mol-1) = 40
...

Note that the mass of sodium carbonate is given in the question as 53
...
It is therefore
known to 3 significant figures and the accuracy given in the answer reflects this
...


Metals form the majority of the elements in the Periodic Table and occur on the left side
of the Table
...
There are a few
mettaloids or semi-metals on the border between metals and non-metals
...

6
...

Physical properties:
• Metals conduct electricity - non-metals do not (the allotrope of carbon, graphite,
is an exception but is not a metallic conductor as its conductivity actually
increases with temperature)
...

• Metals usually have a higher density than non-metals
...


Chemical properties:
• Metals are cationic and non-metals anionic in salts
...

• Some metals react with acid to give hydrogen gas and the metal cation nonmetals do not react in this way
...

N

Cl

H

amount in 100 g

26
...
4

7
...
2
= 1
...
01
1
...
00 ~1
1
...
4
= 1
...
45
1
...
00 ~ 1
1
...
5
= 7
...
08
7
...
98 ~ 4
1
...


The molar mass of H2 is 2 × 1
...
016 g mol-1
...
00 g is:
number of moles =

mass (in g)
molar mass (in g mol

−1

)

=

4
...
016g mol −1

= 1
...
45 = 70
...
The number of moles in 10
...
0g
70
...
141 mol

The reaction is:
H2(g) + Cl2(g) à 2HCl(g)
One mole of H2 reacts with one mole of Cl2 to give two moles of HCl
...
98 – 0
...
84 mol of H2 will be left over
...
Two moles of HCl is produced for every mole of Cl2
...
141
mol of Cl2 available, the amount of HCl that can be formed is 2 × 0
...
282 mol
...
008 (H) + 35
...
458 g mol-1
...
282 mol × 36
...
3 g
9
...
100 M solution is:
number of moles = volume (in L) × concentration (in mol L-1)
=

250
L × 0
...
0250 mol
1000

The formula of sodium carbonate is Na2CO3 so its molar mass is:
molar mass = (2×22
...
01 (C) + 3×16
...
99 g mol-1
The mass of sodium carbonate required is therefore:
mass = number of moles (in mol) × molar mass (g mol-1)
= 0
...
99 g mol-1 = 2
...


The formula of magnesium oxide is MgO so its formula mass is 24
...
00 (O) =
40
...
The number of moles in 2
...
00g
40
...
0496 mol

The number of moles of HCl in 10
...
00 M solution is:
number of moles = volume (in L) × concentration (in mol L-1)
=

10
L × 2
...
0200 mol
1000

The reaction is:
or

MgO(s) + 2HCl(aq) à MgCl2(aq) + H2O(l)

(full)

MgO(s) + 2H+(aq) à Mg2+(aq) + H2O(l)

(ions)

One mole of MgO reacts with two moles of acid
...
0496 mol), (2 × 0
...
0992 mol of acid would be required
...
0200 mol
of acid is available so MgO is in excess and some will be left at the end of the reaction
...
0200 mol of HCl is available, only 0
...
The amount of
MgO remaining at the end of the reaction is therefore 0
...
0100 = 0
...

11
...
01 (N) + 4×1
...
052 g mol-1
The number of moles of hydrazine in 1
...
00 × 102 g
32
...
12 mol

The formula of nitrogen tetraoxide is N2O4 so its molar mass is:
molar mass = (2×14
...
00 (O)) g mol-1 = 92
...
00 × 102 g is therefore:
number of moles (in mol) =

𝐦𝐚𝐬𝐬
𝐦𝐨𝐥𝐚𝐫  𝐦𝐚𝐬𝐬

=

𝟐
...
𝟎𝟐  𝐠  𝐦𝐨𝐥!𝟏

= 2
...
The 2
...
17 = 4
...
Less than this amount is available so
N2H4 is the limiting reagent and controls the amount of product that can be made
...

The amount of N2 formed is limited by the amount of N2H4 available
...
The
number of moles of N2 that will be made from 3
...
12 mol = 4
...
01 = 28
...
The mass of 4
...
68 mol × 28
...


The formula of magnesium hydroxide Mg(OH)2 so its molar mass is:
molar mass = (24
...
00 (O) + 1
...
326 g mol-1
The number of moles of Mg(OH)2 in 0
...
10g
58
...
0017 mol

The chemical equation for the neutralization reaction is:
or

Mg(OH)2(s) + 2HCl(aq) à MgCl2(aq) + 2H2O(l)

(full)

Mg(OH)2(s) + 2H+(aq) à Mg2+(aq) + 2H2O(l)

(ions)

Two moles of acid are neutralized by every one mole of Mg(OH)2
...
0017 mol of
Mg(OH)2 are present in the tablet, 2 × 0
...
0034 mol of acid will be neutralized
...
10 M solution of HCl containing this number of moles is:
volume (in L) =

number of moles(inmol)
concentration(inmol L−1 )

As 1 L = 1000 mL, 0
...


=

0
...
10mol L−1

= 0

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