Limiting Reagent and Yield

To calculate the amount of product to be formed in the reaction, the concepts of limiting reagent and yield are used.

By Sakshi Goel | 28 Oct'18 | 1 K Views |

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Theory

The limiting reactant (or limiting reagent) is the reactant that is completely consumed in a chemical reaction and limits the amount of product.
The reactant in excess is any reactant that occurs in a quantity greater than is required to completely react with the limiting reactant.
Suppose we have a reaction:
$2 H_{2} + O_{2} â†’ 2 H_{2} O$
In this reaction, 2 moles of H₂ combines with 1 mole of O₂. Suppose we are given 2 moles of H₂ and 2 moles of O₂, then this means that O₂ is in excess and H₂ will decide the number of moles of product to be formed. So, H₂ is the limiting reagent and 1 mole of O₂ will be left after the reaction.

The theoretical yield is the amount of product that can be made in a chemical reaction based on the amount of limiting reactant.
The actual yield is the amount of product actually produced by a chemical reaction.
The percent yield is calculated as $\frac{A c t u a l y i e l d}{T h e o r e t i c a l Y i e l d} Ã— 100$

MCQ

Problem:
10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be
(a) 1 mol
(b) 2 mol
(c) 3 mol
(d) 4 mol

Solution:

2H₂ + O₂ $â†’$ 2H₂O
4 g 32 g 2 moles

4g of H₂ require 32 g of O₂.

So, 10 g of H₂ will require = $\frac{32}{4} Ã— 10$ = 80 g

In the question, as 10 g of H₂ is to react with 64 g of O₂

So, O₂ will completely consumed and will act as limiting reagent.
32 g of O₂ gives H₂O = 2 mol

64 g of O₂ gives H₂O = $\frac{2}{32} Ã— 64$ = 4 mol

Option (d) is correct

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Problem:
In the reaction, 2SO₂ + O₂ $â†’$ 2SO₃
when 1 mole of SO₂ and 1 mole of O₂ are made to react to completion
(a) All the oxygen will be consumed
(b) 1.0 mole of SO₃ will be produced
(c) 0.5 mole of SO₂ is remained
(d) All of these

Solution:

2SO₂ + O₂ $â†’$ 2SO₃
2 moles of SO₂ reacts with 1 mole of O₂

So, 1 mol of SO₂reacts with 0.5 mole of O₂.
So, SO₂ will be limiting reagent will formed.

1 mol of SO₃ would be formed

Correct option is (d).

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Problem:
For the reaction $A + 2 B â†’ C$ , 10 moles of A and 16 moles of B will produce
(a) 5 moles C
(b) 4 moles C
(c) 13 moles C
(d) 8 moles C

Solution:

In reaction, 1 mole of A reacts with 2 moles of B.
So, 10 moles of A will react with 20 moles of B.
But 16 moles of B is available.
This means that B is limiting reagent.
2 moles of B produce 1 mole of C.
So, 16 moles of B will produce 8 moles of C
Correct option is (d).

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Problem:
12 g of Mg will react with an acid to give
(a) 1 mole of O₂
(b) 1 mole of H₂
(c) 2 moles of H₂
(d) (1/2) mole of H₂

Solution:

Molar mass of Mg = 24 g/mol
mass of Mg = 12 g
Moles of Mg = 12/24 = 0.5 mol
The reaction of Mg with acid is
$Mg + 2 H^{+} â†’ {Mg}^{2 +} + H_{2}$
1 mole of Mg gives 1 mole of H₂.
So, 0.5 mol of Mg will give 0.5 mole of H₂.
Option (d) is correct.

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Problem:
When 1 mol of ammonia and 1 mol of O₂ are allowed to react then
(a) 1 mol of H₂O is produced
(b) 1 mol of NO is produced
(c) 1 mol of O₂ is consumed
(d) 1 mol of NH₃ is consumed

Solution:

In the reaction, to check the limiting reagent,

	NH₃	O₂	NO	H₂O
In reaction	4 moles	5 moles	4 moles	6 moles
Question	1 mole	$\frac{5}{4} Ã— 1 = 1.25 m o l$	$\frac{4}{4} Ã— 1 = 1 m o l$	$\frac{6}{4} Ã— 1 = 1.5 m o l$

If we react 1 mole of ammonia with 1 mol of oxygen, we can see that oxygen is in excess. So, whole of ammonia will react to give 1 mol of NO and 1.5 mol of water. In the options given, options (b) and (d) are correct.

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Problem:
In the reaction $4 A + 2 B + 3 C â†’ A_{4} B_{2} C_{3}$ , what will be the number of moles of product formed when we started with 1 mole A, 0.6 mol B and 0.72 mol C?
(a) 0.25
(b) 0.24
(c) 0.3
(d) 2.32

Solution:

In the reaction, to check the limiting reagent,

	A	B	C
In reaction	4 moles	2 moles	3 moles
Question	1 mole	$\frac{2}{4} Ã— 1 = 0.5 m o l$	$\frac{3}{4} Ã— 1 = 0.75 m o l$

But we have 0.72 mol C, so C is the limiting reagent.
For 3 moles of C, we get 1 mole of product,
if we have 0.72 mol of C, then moles of product formed will be = $\frac{1}{3} Ã— 0.72 = 0.24 m o l$
Correct option is (b).

Problem: For the reaction,
$4 {NH}_{3} + 5 O_{2} â†’ 4 NO + 6 H_{2} O$

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Problem:
How much SO₂ is produced by burning sulphur in 0.5 mol oxygen?
(a) 16 g
(b) 32 g
(c) 80 g
(d) 64 g

Solution:

The reaction is:
$S + O_{2} â†’ {SO}_{2}$
1 mole of Oxygen gives 1 mole of SO₂.
So, 0.5 mol of oxygen will give 0.5 mol of SO₂.
Molar mass of SO₂ = 64 g/mol
Mass of SO₂ produced = 0.5 x 64 = 32 g
Correct option is (b).

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Problem:
Find the volume of CO required to reduce 0.25 mole ferric oxide as per the given equation
${Fe}_{2} O_{3} + 3 CO â†’ 2 Fe + 3 {CO}_{2}$
(a) 16.8 dm³
(b) 67.2 dm³
(c) 22.4 dm³
(d) 44.8 dm³

Solution:

In the reaction, 1 mole of ferric oxide reacts with 3 moles of CO.
So, 0.25 mole of ferric oxide reacts with moles of CO = 3(0.25) = 0.75 mol
Volume of 1 mole = 22.4 L
Volume of 0.75 mol = 22.4 x 0.75 = 16.8 L
1 L = 1 dm³
So, 16.8 L = 16.8 dm³
Correct option is (a).

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Problem:
The mass of BaCO₃ produced when excess CO₂ is bubbled through a solution of 0.205 mol Ba(OH)₂ is
(a) 81 g
(b) 20.25 g
(c) 40.5 g
(d) 162 g

Solution:

The reaction is:
$Ba (OH)_{2} + {CO}_{2} â†’ {BaCO}_{3} + H_{2} O$
CO₂ is in excess, so Ba(OH)₂ is the limiting reagent.
Moles of Ba(OH)₂reacted = 0.205 mol
Moles of BaCO₃ formed = Moles of Ba(OH)₂reacted = 0.205 mol
Molar mass of BaCO₃ = 197 g/mol
Mass of BaCO₃ formed = 0.205 x 197 = 40.4 g
Closest option is (c).

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Problem:
For the reaction $A + 2 B â†’ C$ , 5 moles of A and 8 moles of B will produce
(a) 5 moles C
(b) 4 moles C
(c) 8 moles C
(d) 13 moles C

Solution:

In reaction, 1 mole of A reacts with 2 moles of B.
So, 5 moles of A will react with 10 moles of B.
But 8 moles of B is available.
This means that B is limiting reagent.
2 moles of B produce 1 mole of C.
So, 8 moles of B will produce 4 moles of C
Correct option is (b).

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Problem:
Assuming fully decomposed, the volume of CO₂ released at STP on heating 9.85 g BaCO₃ will be
(a) 1.12 L
(b) 0.84 L
(c) 2.24 L
(d) 4.06 L

Solution:

The reaction is:
${BaCO}_{3} â†’ BaO + {CO}_{2}$
molar mass of BaCO₃ = 197 g/mol
Mass of BaCO₃ decomposed = 9.85 g
Moles of BaCO₃ decomposed = 9.85/197 = 0.05 mol
1 mole of BaCO₃ decomposes to give 1 mole of CO₂.
0.05 mole of BaCO₃ decomposes to give 0.05 mole of CO₂.
At STP, Volume of 1 mole of gas = 22.4 L
Volume of 0.05 mol of gas = 22.4 x 0.05 = 1.12 L
Correct option is (a).

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Problem:
Find the volume of CO₂ at NTP obtained on heating 10 g of 90 % pure limestone
(a) 22.4 L
(b) 20.16 L
(c) 2.24 L
(d) 2.016 L

Solution:

Limestone is CaCO₃. So, the reaction is:
${CaCO}_{3} â†’ CaO + {CO}_{2}$
Limestone is 90% pure. So actual mass of CaCO₃ reacting = 90% of 10 g
= (90/100) x 10 = 9 g
Molar mass of CaCO₃ = 100 g/mol
Moles of CaCO₃ reacting = mass/molar mass = 9/100 = 0.09 mol
According to reaction, 1 mole of CaCO₃ reacts to give 1 mole of CO₂.
So, 0.09 mol of CaCO₃ reacts to give 0.09 mole of CO₂.
At NTP, Volume of 1 mole of gas = 22.4 L
Volume of 0.09 mol of gas = 22.4 x 0.09 = 2.016 L
Correct option is (d).

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Problem:
2.76 g of silver carbonate on being strongly heated at quite high temperature yields a residue weighing
(a) 2.64 g
(b) 2.48 g
(c) 2.32 g
(d) 2.16 g

Solution:

The reaction of heating silver carbonate is:
${Ag}_{2} {CO}_{3} â†’ 2 Ag (s) + {CO}_{2} (g) + \frac{1}{2} O_{2} (g)$
In the reaction, only Ag is residue.
Molar mass of Ag₂CO₃ = 276 g/mol
Moles of Ag₂CO₃ in 2.76 g = mass/molar mass = 2.76/276 = 0.01 mol
1 mole of Ag₂CO₃ gives 2 moles of Ag
So, 0.01 mol of Ag₂CO₃ gives 0.02 moles of Ag
Molar mass of Ag = 108 g/mol
Mass of 0.02 mol of Ag = 108 x 0.02 = 2.16 g
Mass of residue = 2.16 g
Option (d) is correct.