Y 10 Chem Solving Moles Problems 1

[ Relative atomic mass:

Relative Atomic Mass

FIRST STEP:

Mass

Calculate the number of atoms in
1.28 g of copper?

Find the

Cu = 64; N_A = 6 x 10²³mol^-1]

Number of MOLES

using 1.28g

mol

Relative Molecular Mass

[ Relative atomic mass:

Mass

FIRST STEP:

Calculate the number of

in 4.4g of vitamin C, C₆H₈O₆

Find the

H = 1; C = 12; O = 16;
N_A = 6 x 10²³mol^-1]

Number of MOLES

molecules

using 4.4g

mol

use 0.6 g to find the number of moles of NH₃

[ Relative atomic mass:

Use N_Ato find the number of atoms of NH₃

in 0.6g of NH₃gas evolved during an experiment

FIRST STEP?

What is the

Calculate the number of

N = 14 ; H = 1; O = 16;
N_A = 6.02 x 10²³mol^-1]

atoms

Relative Molecular Mass

[ Relative atomic mass:

Mass

FIRST STEP:

in 0.6g of NH₃gas evolved during an experiment

Calculate the number of

use 0.6 g to Find the

N = 14 ; H = 1; O = 16;
N_A = 6.02 x 10²³mol^-1]

atoms

Number of MOLES

0.035

mol

use RMM to find the number of atoms of H₂O₂

use 10 g to find the number of moles of H₂O₂

[ Relative atomic mass:

What is the first step?

is used, find the mass of oxygen released?

FIRST STEP:

If 10 g of hydrogen peroxide

Hydrogen peroxide decomposes as below

2H₂O₂

H = 1; O = 16;]

2H₂O + O₂

Mass

Relative Molecular Mass

FIRST STEP:

[ Relative atomic mass:

is used, find the mass of oxygen released?

If 10 g of hydrogen peroxide

Hydrogen peroxide decomposes as below

use 10g to Find the

2H₂O₂

H = 1; O = 16;]

2H₂O + O₂

Number of MOLES

0.147

mol

The 2nd STEP

mole=

mole =

mole=

22.4 dm³(STP)

No. atom

24dm³(room t)

vol

N_A

mole=

RAM/RMM/RFM

the problems

use the correct

FORMULA

mass

No. molecule

N_A

to solve

[ Relative atomic mass:

FIRST STEP:

SECOND STEP:

Calculate the number of atoms in
1.28 g of copper?

use 1.28g to Find the

USE THE FORMULA

Cu = 64; N_A = 6 x 10²³mol^-1]

Mole =

No of Atoms

Number of MOLES

[ Relative atomic mass:

What is the

number of mole =number of atoms x N_A

number of mole = number of atoms ÷ N_A

Calculate the number of atoms in
1.28 g of copper?

use the formula

SECOND STEP ?

Cu = 64; N_A = 6 x 10²³mol^-1]

Number of C₆H₈O₆ Molecules=

[ Relative atomic mass:

SECOND STEP:

From the first step, the number of moles of C₆H₈O₆ is

Number of MOLES

0.025 mol

10.

Calculate the number of

in 4.4g of vitamin C, C₆H₈O₆

Use the formula

H = 1; C = 12; O = 16;
N_A = 6 x 10²³mol^-1]

number of molecules

molecules

6X10²³mol^-1

First: use 0.6g to find the number of moles of NH₃,
SECOND: use formula of mole=No. of atoms ÷ N_A

[ Relative atomic mass:

11.

first: Use numbers of atoms x N_A, then
Second step: find the number of moles of NH₃

What is the step of solving this?

in 0.6g of NH₃gas evolved during an experiment

Calculate the number of

N = 14 ; H = 1; O = 16;
N_A = 6.02 x 10²³mol^-1]

atoms

First: use 3.4g to find the number of moles of NH₃,
SECOND: use formula of mole=No. of atoms ÷ N_A

[ Relative atomic mass:

first: Use the formula of mole x N_A, then
Second step: find the number of moles of NH₃

12.

What is the step of solving this?

in 3.4g of NH₃gas evolved during an experiment

Calculate the number of

N = 14 ; H = 1;
N_A = 6 x 10²³mol^-1]

atoms

Step 1:

Step 2:

Now use the

[ Relative atomic mass:

13.

in 3.4g of NH₃gas evolved during an experiment

Calculate the number of

No. of NH₃

No. of Moles of NH₃=

ANSWER

molecules

to find the no. of

N = 14 ; H = 1;
N_A = 6 x 10²³mol^-1]

N_A

atoms

mol^-1

NH₃

mol^-1

molecules

X 10²³

Step 2:

Step 1:

Step 3:

[ Relative atomic mass:

14.

we want to find number of

in 1 molecule of NH₃

how many atoms

in 3.4g of NH₃gas evolved during an experiment

No. of NH₃

No. of Moles of NH₃=

Calculate the number of

No. of NH₃

molecules =

ATOMS

3.4

0.2 mol^-1

N = 14 ; H = 1;
N_A = 6 x 10²³mol^-1]

0.2

Atoms

atoms

6 X 10²³

mol^-1

0.2 mol^-1

X10²³

of NH₃

atoms

x 6 x10²³

Step 3:

Step 2:

Step 1:

[ Relative atomic mass:

15.

of hydrated copper (II) sulphate CuSO₄.5H₂O

Calculate the mass of

Relative Formula Mass

No. of Moles of CuSO₄.5H₂O=

Mass

of CuSO₄.5H₂O =

= mole x RFM

O = 16 ; H = 1;
S = 32; Cu = 64]

0.3

mol

no need to find

mol

given

Step 3:

Step 2:

Step 1:

[ Relative atomic mass:

16.

of Sulphur trioxide SO₃

Calculate the mass of

Relative Formula Mass

No. of Moles of SO₃=

Mass

of SO₃ =

= mole x RFM

O = 16 ; S = 32 ]

0.5

mol

given

Step 2:

Step 1:

Step 3:

[ Relative atomic mass:

17.

of Sulphur trioxide SO₃

Calculate the number of atoms

1 molecule of SO₃

No. of atoms of
1.5 M of SO₃

No. of Moles of SO₃=

O = 16 ; S = 32;
N_A=6.02 x 10²³ ]

atoms of S, O,O,O

mole x Avogadro

in 1.5

mol

given

x10²³

Step 2:

Step 1:

Step 3:

[ Relative atomic mass:

18.

which has 3.01 x 10²³

Calculate the mass

RFM of CO₂

Mass of CO₂

No. of Moles of CO₂=

O = 16 ; C = 12;
N_A=6.02 x 10²³ ]

molecules

of carbon dioxide CO₂

No of molecules

mole x RFM

N_A

in STP

mol

[ Relative atomic mass:

From the equation, 1 mol of Oxygen O₂ used

19.

when S + O₂→ SO₂and the O₂ used is 67.2 dm³ at STP?

produce

Calculate the

S + O₂

mole of SO₂

volume

O = 16 ; S = 32;
N_A=6.02 x 10²³ ]

→

of sulphur dioxide SO₂

SO₂

Moles:

molar gas volume=22.4 dm³ STP

Volume:

No. of moles

=Vol ÷ 22.4

when S + O₂→ SO₂and the O₂ used is 67.2 dm³ at STP?

20.

Calculate the

[ Relative atomic mass:

S + O₂

67.2

volume

dm³

→

SO₂

of sulphur dioxide SO₂

O = 16 ; S = 32;
N_A=6.02 x 10²³ ]

Volume of SO₂

dm³

Step 2:

Step 1:

Step 3:

21.

at STP

Calculate the

Find the RMM

Divide mass of C₂H₆by RMM

use the mole x molar gas volume=vol

[ Relative atomic mass:

H = 1 ; C = 12;
molar gas volume=22.4 dm³ STP ]

volume

of 0.75g of ethane gas C₂H₆

Step 2:

Step 1:

Step 3:

22.

at STP

Calculate the

No. of mole of C₂H₆

Volume of C₂H₆

RMM of C₂H₆=

[ Relative atomic mass:

H = 1 ; C = 12;
molar gas volume=22.4 dm³ STP ]

volume

of 0.75g of ethane gas C₂H₆

dm³

dm³ x

mol

^mol

Step 2:

Step 1:

Step 3:

23.

at room temperature

Calculate the

Get mole=Divide vol of C₃H₆by Molar gas vol

use the mole x RMM =mass

Find the RMM

mass

H = 1 ; C = 12;
molar gas volume=24 dm³ R.T. ]

of 6.72 dm³of propene gas C₃H₆

[ Relative atomic mass:

Step 2:

Step 1:

Step 3:

24.

at room temperature

Calculate the

No. of mole of C₃H₆

Mass of C₃H₆

RMM of C₃H₆=

mass

H = 1 ; C = 12;
molar gas volume=24 dm³ R.T. ]

of 6.72 dm³of propene gas C₃H₆

[ Relative atomic mass:

^g x

mol

^mol

_Moles:

Step 1:

No. of
Moles

when 0.5 dm³ of mole of

25.

Calculate the

Study the balanced equation

0.5

2H₂O₂ → 2H₂O + O₂_.

volume

of oxygen gas O₂produced

H₂O₂

decomposed

1 x 0.5 ÷ 2

To find Volume of O₂ :

Step 2:

Volume:

when 0.5 dm³ of mole of

26.

Calculate the

given

2000÷4

the volume

0.5

volume

2000 cm³

500

2H₂O₂ → 2H₂O + O₂_.

of oxygen gas O₂produced

H₂O₂

÷ 4

decomposed

24000 cm³

no. of
moles

mol

Moles:

No. of
moles

when 200 cm³ of 0.5 mol of

27.

2H₂O₂ → 2H₂O + O₂_.

Calculate the

[ Relative atomic mass:

2H₂O₂

0.5

→

volume

2H₂O

+ O₂

of oxygen gas O₂produced

H₂O₂

O = 16 ; H = 1;
24dm³ at room tem ]

decomposed

when 200 cm³ of 0.5 mol of

28.

2H₂O₂ → 2H₂O + O₂_.

Volume

No. of
moles

Moles

Calculate the

2000 cm³

[ Relative atomic mass:

2H₂O₂

0.5

→

volume

2H₂O

+ O₂

0.25

of oxygen gas O₂produced

H₂O₂

O = 16 ; H = 1;
24dm³ at room tem ]

cm³

decomposed

when 200 cm³ of 0.5 mol of

29.

2H₂O₂ → 2H₂O + O₂_.

Volume

÷10

No. of
moles

Moles

Calculate the

2000 cm³

[ Relative atomic mass:

2H₂O₂

0.5

cm³

→

volume

2H₂O

24000

+ O₂

0.25

of oxygen gas O₂produced

H₂O₂

O = 16 ; H = 1;
24dm³ at room tem ]

cm³

decomposed

Volume of O₂

cm₃

cm³