STOICHIOMETRY

MOLE CONCEPT

1

The molar mass of H₂SO₄ is ...

Ar H=1, S=32, O=16

H₂SO₄

Mr = 2 x Ar H + 1 x Ar S + 4 x Ar O

Mr = 2(1)  + 32 + 4(16) =98

1

Find the Mr of hydrated copper (II) sulfate,

CuSO₄.₅ H₂O

Ar Cu =63.5, S=32, O=16,H=1

Mr of CuSO₄.₅ H₂O

Mr CuSO₄  +  5 x Mr H₂O

Mr CuSO₄ = 63.5 + 32 + 4(16) =159.5

Mr H₂O = 2(1)  + 16 = 18

Mr of CuSO₄.₅ H₂O = Mr CuSO₄  +  5 x Mr H₂O

 Mr of CuSO₄.₅ H₂O = 159.5 + 5(18) = 249.5

2

Find the mass of 2 mole NH₃ if Ar N =14, H=1 

Mr NH₃ = 1 x Ar N + 3 x Ar H 

Mr NH₃ = 14 + 3 (1) = 17

Mass NH₃  = mol x Mr

Mass NH₃  = 2 x 17 = 34 gram

2

Find the volume of 1.5 mol CO₂(g) at RTP.

Vm at RTP = 24.4 L

Volume of CO₂ at RTP = mol x Vm

V CO₂ = 1.5 x 24.4 L = 36.6 L

3

Find the mass of 11.2 mL of CH₄(_g) at STP. 

Ar C=12, H=1

Vol CH₄ = 11.2 : 1000 = 0.0112 L

Mol CH₄ = V : Vm = 0.0112 L : 22.4 L = 0.0005 mol

Mr CH₄  = 12 + 4(1) = 16

Mass CH₄  = mol x Mr = 0.0005 x 16 = 0.008 gram

 

STOICHIOMETRY

1

The balanced equation for the following reaction : 

Na₂CO₃ + HCl -> NaCl + H₂O + CO₂

is...

Na₂CO₃ + 2 HCl -> 2 NaCl + H₂O + CO₂

1

Given the following reaction :

Mg(s) + 2 HCl (aq) -> MgCl₂ (aq)  + H₂ (g)

Find the mole of H₂ formed when 2 mol of Mg is used in the reaction.

Mole ratio of Mg : HCl : MgCl₂ : H₂ = 1 : 2: 1 : 1

Mole of Mg = 2 mol

Mole of H₂ = 2 mol

2

Given the following reaction :

2 SO₂ (g) + O₂ (g) -> 2 SO₃ (g)

Find the mass of SO₃ formed when 6.4 grams of  SO₂ are used in the reaction!

Ar S= 32, O=16

Mole ratio SO₂ : O₂ : SO₃ = 2: 1: 2

Mr SO₂ = 1(32) + 2(16) =64

Mole of SO₂ = mass : Mr = 6.4 : 64 = 0.1 mol

Mole of SO₃ = 0.1 mol

Mr SO₃ = 1(32) + 3(16) = 80

Mass SO₃ = mole x Mr = 0.1 x 80 = 8 gram.

2

Given the following reaction :

Mg (s) + 2 HCl (aq) -> MgCl₂ (aq) + H₂ (g)

Find the volume of H₂ formed when 0.4 mole of HCl is reacted at P, T.

The 1 ml of  N₂ mass is 7 mg at P, T.

Ar N=14

Mole of HCl = 0.4 mol

Mole of H₂ = 0.4 : 2 = 0.2 mol = 200 mmol

At P, T (compare the gas if H₂ and N₂)

Mr N₂ = 2(14) =28

Mole of N₂ = 7: 28 = 0.25 mmol 

V₁ / n₁ = V₂ /n₂

1 ml/ 0.25 mmol = VH₂ /200 mmol

V H2 = 800 ml

mole concept 2

1

Find the mole of CH₃COOH dissolved in 100 ml solution of CH₃COOH 0.04 M.

n = C x V

n = 0.04 x 0.1 = 0.004 mol

2

Find the mass of glucose dissolved in 200 ml of the glucose solution  0.06 M. Mr glucose = 180

mole of glucose = C x V = 0.06 x 0.2 = 0.012 mol

Mass of glucose = mole x Mr = 0.012 x 180 = 2.16 grams.

2

Find the mass of 3.01 x 10 ²⁰ molecules of NH₃ if Mr of NH₃ = 17.

NA = 6.02 x10 ²³

Mole of NH₃ = (3.01 x 10 ²⁰ ) : ( 6.02 x10 ²³ )= 5 x10^-4

Mass = mole x Mr = 0.0005 x 17 = 0.0085 grams.

Percent Yield

1

Given a percent yield of a reaction is 75%, and the mass of the actual yield is 8 grams. Find the mass of the theoretical yield!

Mass of theoretical yield = Mass of actual yield: the percent yield 

= 8: 0.75 = 10.67 grams.

2

Given a 100 ml AgNO₃ solution 0.05 M, it is reacted with HCl via the following reaction :

AgNO₃ + HCl -> AgCl +HNO₃

The mass of AgCl formed is 0.6567 grams. Find the percent yield if the Mr AgCl = 143.5

Mole of AgNO₃ = C x V = 0.05 x 0.1 = 0.005 mol

Mole of AgCl = 0.005 mol

Theoretical yield of AgCl = mole x Mr = 0.005 x 145.5 = 0.7175 grams

Percent yield = ( 0.6567 : 0.7175) x 100% = 91.52 %

LIMITING REACTANT

1

A 7 mol of CH₄ was combusted with 10 mol of O₂ via the following reaction :

CH₄ + 2 O₂ -> CO₂ + 2 H₂O

The limiting reactant is ...

CH₄ = 7 :1 = 7

O₂ = 10 : 2 = 5

O₂ < CH₄

The LR is O₂

2

A 7 mol of CH₄ was combusted with 10 mol of O₂ via the following reaction :

CH₄ + 2 O₂ -> CO₂ + 2 H₂O

The final mole of CH₄ is ...

Initial mole of CH₄ = 7 mol

LR = O₂

Mole of CH₄ used in the reaction = 5 mol

Final mole of O₂ = 7 mol - 5 mol = 2 mol

2

A 7 mol of CH₄ was combusted with 10 mol of O₂ via the following reaction :

CH₄ + 2 O₂ -> CO₂ + 2 H₂O

Find the mass of CO₂ formed! Mr CO₂ = 44

LR = O₂

The mole of CO₂ formed = 10 mol : 2 = 5 mol

Mass CO₂ = mol x Mr = 5 x 44 = 220 gram.