The reaction between red lead and hydrochloric acid is given below, Pb 3 O 4 + 8 HCl → 3 PbCl 2 + 4 H 2 O + Cl 2 . Calculate the mass of lead chloride formed by the action of 6 . 85 g of red lead.

The reaction between red lead and hydrochloric acid is given below, Pb 3 O 4 + 8 HCl → 3 PbCl 2 + 4 H 2 O + Cl 2 . The molar mass of red lead, Pb 3 O 4 = 3 × 207 . 2 + 4 × 16 = 685 . 6 g mol - 1 . The number of moles of red lead, ⇒ given mass molar mass = 0 . 01 mol . In the above reaction, one mole of red lead forms three moles of lead chloride. Therefore, 0 . 01 mol of red lead forms 3 × 0 . 01 = 0 . 03 mol of lead chloride. The mass of lead chloride will be, 0 . 03 = x g 207 . 2 + 3 × 35 . 45 ⇒ x g = 313 . 55 × 0 . 03 = 9 . 407 g . Thus, the mass of lead chloride is 9 . 407 g .

The reaction between red lead and hydrochloric acid is given below, Pb 3 O 4 + 8 HCl → 3 PbCl 2 + 4 H 2 O + Cl 2 . 6 . 856 g of red lead involved in the reaction. Calculate the mass of chlorine.

The reaction between red lead and hydrochloric acid is given below, Pb 3 O 4 + 8 HCl → 3 PbCl 2 + 4 H 2 O + Cl 2 . The molar mass of red lead, Pb 3 O 4 = 3 × 207 . 2 + 4 × 16 = 685 . 6 g mol - 1 . The number of moles of red lead, ⇒ given mass molar mass = 0 . 01 mol . In the above reaction, one mole of red lead forms one mole of chlorine. Therefore, 0 . 01 mol of red lead forms 0 . 01 mol of chlorine. The mass of lead chloride will be, 0 . 01 = x g 71 ⇒ x g = 71 × 0 . 01 = 0 . 71 g . Thus, the mass of chlorine is 0 . 71 g .

The reaction between red lead and hydrochloric acid is given below, Pb 3 O 4 + 8 HCl → 3 PbCl 2 + 4 H 2 O + Cl 2 . Calculate the volume of chlorine evolved at STP when 6 . 856 g of red lead involved in the reaction.

The reaction between red lead and hydrochloric acid is given below, Pb 3 O 4 + 8 HCl → 3 PbCl 2 + 4 H 2 O + Cl 2 . The molar mass of red lead, Pb 3 O 4 = 3 × 207 . 2 + 4 × 16 = 685 . 6 g mol - 1 . The number of moles of red lead, ⇒ given mass molar mass = 0 . 01 mol . In the above reaction, one mole of red lead forms one mole of chlorine. Therefore, 0 . 01 mol of the red lead produces 0 . 01 mol of the chlorine gas. So, the volume of the chlorine will be, 0 . 01 = x L 22 . 4 L ⇒ x L = 0 . 01 × 22 . 4 = 0 . 224 L . Thus, the volume of the chlorine is 0 . 224 L .

1 . 56 g of sodium peroxide reacts with water according to the following equation, 2 Na 2 O 2 + 2 H 2 O → 4 NaOH + O 2 . Calculate mass of sodium hydroxide formed.

Given, Molar Mass of Na 2 O 2 = 2 × 23 + 2 × 16 = 78 g mol - 1 . Mass of two moles of Na 2 O 2 = 1 . 56 g . Two moles of sodium peroxide produces four moles of sodium hydroxide. So, 1 . 56 g Na 2 O 2 gives = 4 × 40 2 × 78 × 1 . 56 = 1 . 6 g of sodium hydroxide.

If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of B ? The gases A and B are under the same conditions of temperature and pressure. Name the law on which the problem is based.

According to Avogadro’s law, equal volumes of all gases under similar conditions of temperature and pressure contain an equal number of molecules. So 150 cc of B will also contain X molecules. Hence, the number of molecules present in 75 cc of B will contain X 2 molecules.

Calculate the number of moles in 12 g of oxygen gas. [ O = 16 ]

Molecular mass = 16 × 2 = 32 g 32 g of oxygen gas → 1 mole 1 g of oxygen gas → 1 32 mole Given that 12 g of oxygen gas Number of moles = 12 32 = 0 . 375 mole

EASY

10th ICSE

IMPORTANT

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$66 g$ ammonium sulphate is produced by the action of ammonia on sulphuric acid. Write a balanced equation and calculate the mass of acid required.

Important Questions on Mole Concept and Stoichiometry

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 4

The reaction between red lead and hydrochloric acid is given below,

${Pb}_{3} O_{4} + 8 HCl \to 3 {PbCl}_{2} + 4 H_{2} O + {Cl}_{2}$ .

Calculate the mass of lead chloride formed by the action of $6.85 g$ of red lead.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 4

The reaction between red lead and hydrochloric acid is given below,

${Pb}_{3} O_{4} + 8 HCl \to 3 {PbCl}_{2} + 4 H_{2} O + {Cl}_{2}$ .

$6.856 g$ of red lead involved in the reaction. Calculate the mass of chlorine.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 4

The reaction between red lead and hydrochloric acid is given below,

${Pb}_{3} O_{4} + 8 HCl \to 3 {PbCl}_{2} + 4 H_{2} O + {Cl}_{2}$ .

Calculate the volume of chlorine evolved at STP when $6.856 g$ of red lead involved in the reaction.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 5

Find the mass of ${KNO}_{3}$ required to produce $126 kg$ of nitric acid. Find whether a larger or smaller mass of ${NaNO}_{3}$ is required for the same purpose.
${KNO}_{3} + H_{2} {SO}_{4} \to {KHSO}_{4} + {HNO}_{3} . {NaNO}_{3} + H_{2} {SO}_{4} \to {NaHSO}_{4} + {HNO}_{3} .$

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 6

Pure calcium carbonate and dilute hydrochloric acid are reacted and $2$ litres of carbon dioxide was collected at $27 ° C$ and normal pressure.

${CaCO}_{3} + 2 HCl \to {CaCl}_{2} + H_{2} O + {CO}_{2}$ .

Calculate the mass of salt required.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 6

Pure calcium carbonate and dilute hydrochloric acid are reacted and $2$ litres of carbon dioxide was collected at $27 ° C$ and normal pressure.

${CaCO}_{3} + 2 HCl \to {CaCl}_{2} + H_{2} O + {CO}_{2}$ .

Calculate the mass of the acid required.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 7

Calculate the mass and volume of oxygen at STP, which will be evolved on electrolysis of one mole (

18 g

) of water.

EASY

10th ICSE

IMPORTANT

Selina Icse Concise Chemistry For Class 10>Chapter 5 - Mole Concept and Stoichiometry>EXERCISE-5D>Q 8

1.56 g

of sodium peroxide reacts with water according to the following equation,

2 {Na}_{2} O_{2} + 2 H_{2} O \to 4 NaOH + O_{2}

. Calculate mass of sodium hydroxide formed.

66 g ammonium sulphate is produced by the action of ammonia on sulphuric acid. Write a balanced equation and calculate the mass of acid required.

Important Questions on Mole Concept and Stoichiometry

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$66 g$ ammonium sulphate is produced by the action of ammonia on sulphuric acid. Write a balanced equation and calculate the mass of acid required.