67.2 L of hydrogen combines with 44.8 L of nitrogen to form ammonia under specific conditions as N 2 g + 3 H 2 g → 2 N H 3 g Calculate the volume of ammonia produced. What is the other substance, if any that remains in the resultant mixture?

N 2 + 3 H 2 → 2 N H 3 1 v o l . 3 v o l . 2 v o l . 22 .4 L 67 .2 L 44 .8 L We know that one mole of any gas at STP occupies 22 . 4 L of gas. So, 1 volume of any gas will have 22 . 4 L of the gas. It is called the molar volume. Hence, 1 volume of nitrogen N 2 occupies 22 . 4 L 3 volumes of hydrogen, H 2 occupies 3 × 22 . 4 = 67 . 2 L 2 volumes of ammonia, NH 3 occupies 2 × 22 . 4 = 44 . 8 L According to the given balanced equation we know that: 3 volume of H 2 will give 2 volume of N H 3 ∴ 67 .2 L of H 2 will give = 2 × 67.2 3 = 44 .8 L of N H 3 At the same time, 3 volume of H 2 will react with 1 volume of N 2 . ∴ 67.2 L of H 2 will react = 1 3 × 67.2 = 22.4 L of N 2 But in the question it is given that the amount of nitrogen available is 44 . 8 L Therefore, 44.8 - 22.4 = 22.4 L of N 2 will be left unreacted in the mixture.

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation. P + 5 H N O 3 c o n c . → H 3 P O 4 + H 2 O + 5 N O 2 If 9.3 g of phosphorus was used in the reaction, Calculate the number of moles of phosphorus taken. ( H = 1 , N = 14 , P = 31 , O = 16 )

Given: The molar mass of Phosphorus = 31 g / mol The number of moles can be found using the formula, Number of moles = mass of the given sample Molar mass of the element Number of moles of P = 9.3 31 = 0.3 moles.

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation. P + 5 H N O 3 c o n c . → H 3 P O 4 + H 2 O + 5 N O 2 If 9.3 g of phosphorus was used in the reaction, Calculate the mass of phosphoric acid formed. ( H = 1 , N = 14 , P = 31 , O = 16 ) .

Given: Atomic masses of the following elements ( H = 1 , N = 14 , P = 31 , O = 16 ) Molecular mass of phosphoric acid( H 3 PO 4 ) = 3 ( 1 ) + 31 + 4 ( 16 ) = 98 If 31 g of phosphorus produces 98 g of phosphoric acid Then the amount of phosphoric acid produced by 9 . 3 g of phosphorus= 98 31 × 9 . 3 = 29 . 3 g .

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation. P + 5 H N O 3 c o n c . → H 3 P O 4 + H 2 O + 5 N O 2 If 9.3 g of phosphorus was used in the reaction, calculate the volume of nitrogen dioxide produced at S T P . H = 1 , N = 14 , P = 31 , O = 16

Given: The atomic masses of the elements: H = 1 , P = 31 , N = 14 , O = 16 We know that one volume of any gas at STP occupies 22 . 4 L From the equation it is understood that 5 volumes of nitrogen dioxide is formed. Keeping that in mind, 1 volume of phosphorus produces 5 ( 22 . 4 ) L = 112 L So, the volume of nitrogen dioxide produced by 9 . 3 g of phosphorus = 112 31 × 9 . 3 = 33 . 6 L .

Ethane burns in oxygen to form C O 2 and H 2 O according to the equation, 2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O If 1 250 c c of oxygen is burnt with 300 c c of ethane. Calculate the volume of C O 2 formed.

2 C 2 H 6 + 7 O 2 → 4 C O 2 + 6 H 2 O According to the above reaction where ethane burns in the presence of oxygen, 2 volumes of ethane reacts with oxygen to give 4 volumes of carbon dioxide. This means the volume of carbon dioxide formed is two times the volume of ethane. So the volume of carbon dioxide formed from 300 cc of ethane would be two times the amount of ethane = ( 300 × 2 ) cc Therefore, the amount of carbon dioxide formed from 300 cc of ethane is 600 cc .

E M B I B E

All in One - ICSE Chemistry>Chapter 5 - Mole Concept and Stoichiometry>ARCHIVES*(Last 7 Years)>Q 17

HARD

10th ICSE

IMPORTANT

Earn 100

$O_{2}$ is evolved by heating ${K ClO}_{3}$ using ${M n O}_{2}$ as a catalyst
$2 KC {lO}_{3} \overset{{MnO}_{2}}{\to} 2 KCl + 3 O_{2}$
(i)Calculate the mass of ${K ClO}_{3}$ required to produce $6.72 L$ of $O_{2}$ at $S T P$ . $(K = 39, C l = 35.5 a n d O = 16)$

(ii) Calculate the number of moles of oxygen present in the $6.72 L$ volume and also the number of molecules.

Important Questions on Mole Concept and Stoichiometry

MEDIUM

10th ICSE

IMPORTANT

All in One - ICSE Chemistry>Chapter 5 - Mole Concept and Stoichiometry>ARCHIVES*(Last 7 Years)>Q 17

Calculate the volume occupied by

0.01

mole of

{C O}_{2}

at

S T P

.

MEDIUM

10th ICSE

IMPORTANT

All in One - ICSE Chemistry>Chapter 5 - Mole Concept and Stoichiometry>ARCHIVES*(Last 7 Years)>Q 18

The vapour density of carbon dioxide

(C = 12 a n d O = 16)

is

HARD

10th ICSE

IMPORTANT

All in One - ICSE Chemistry>Chapter 5 - Mole Concept and Stoichiometry>ARCHIVES*(Last 7 Years)>Q 19

67.2 L

of hydrogen combines with

44.8 L

of nitrogen to form ammonia under specific conditions as

N_{2} (g) + 3 H_{2} (g) \to 2 {N H}_{3} (g)

Calculate the volume of ammonia produced. What is the other substance, if any that remains in the resultant mixture?

MEDIUM

10th ICSE

IMPORTANT

All in One - ICSE Chemistry>Chapter 5 - Mole Concept and Stoichiometry>ARCHIVES*(Last 7 Years)>Q 20

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation.
$P + 5 H N O_{3} (c o n c .) \to H_{3} {P O}_{4} + H_{2} O + 5 {N O}_{2}$
If $9.3 g$ of phosphorus was used in the reaction,

Calculate the number of moles of phosphorus taken.

$(H = 1, N = 14, P = 31, O = 16)$

MEDIUM

10th ICSE

IMPORTANT

All in One - ICSE Chemistry>Chapter 5 - Mole Concept and Stoichiometry>ARCHIVES*(Last 7 Years)>Q 20

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation.
$P + 5 H N O_{3} (c o n c .) \to H_{3} {P O}_{4} + H_{2} O + 5 {N O}_{2}$
If $9.3 g$ of phosphorus was used in the reaction,