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#stoichiometry

170 public questions tagged with this topic.

CHEMISTRY

What is the mole fraction of NaCl in a solution containing 5.85 g of NaCl and 90 g of water? (Molar masses: NaCl = 58.5

Given: What is the mole fraction of NaCl in a solution containing 5.85 g of NaCl and 90 g of water? (Molar masses: NaCl = 58.5 g/mol, H₂O = 18 g/mol) These values define the system as per NCERT data. Formula: Moles of NaCl = 5.85 / 58.5 = 0.1 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles of H₂O = 90 / 18 = 5 mol. Total moles = 0.1 + 5 = 5.1. Mole fraction = 0.1 / 5.1 ≈ 0.0196. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

In the Carius method, 0.25 g of a compound gave 0.574 g of AgBr . What is the percentage of bromine? (Atomic masses: Ag

Given: In the Carius method, 0.25 g of a compound gave 0.574 g of AgBr . What is the percentage of bromine? (Atomic masses: Ag = 108, Br = 80) These values define the system as per NCERT data. Formula: Molar mass of AgBr = 108 + 80 = 188 g/mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Mass of Br = 80 × 0.574/188 = 0.244 g. Percentage = 0.244 × 100/0.25 = 97.6% . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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PHYSICS

A gas occupies 5.6 litres at STP. How many moles of the gas are present? (Molar volume at STP = 22.4 litres)

Given: A gas occupies 5.6 litres at STP. How many moles of the gas are present? (Molar volume at STP = 22.4 litres) Formula: Number of moles (μ) = fracVolumeMolar volume. Substitution & Calculation: μ = 5.6/22.4 = 0.25 mol. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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CHEMISTRY

How many molecules are in 12 g of CH₃OH? (Molar mass = 32 g/mol, Avogadro number = 6.022 × 10²³)

Given: How many molecules are in 12 g of CH₃OH? (Molar mass = 32 g/mol, Avogadro number = 6.022 × 10²³) These values define the system as per NCERT data. Formula: Moles = 12 / 32 = 0.375 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Molecules = 0.375 × 6.022 × 10²³ ≈ 2.258 × 10²³. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

How many molecules are in 22 g of CO₂? (Molar mass = 44 g/mol, Avogadro number = 6.022 × 10²³)

Given: How many molecules are in 22 g of CO₂? (Molar mass = 44 g/mol, Avogadro number = 6.022 × 10²³) These values define the system as per NCERT data. Formula: Moles = 22 / 44 = 0.5 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Molecules = 0.5 × 6.022 × 10²³ = 3.011 × 10²³. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

What is the mass of 2 × 10²³ molecules of N₂O? (Molar mass = 44 g/mol, Avogadro number = 6 × 10²³)

Given: What is the mass of 2 × 10²³ molecules of N₂O? (Molar mass = 44 g/mol, Avogadro number = 6 × 10²³) Formula: Moles = 2 × 10²³ / 6 × 10²³ ≈ 0.3333 mol. Substitution & Calculation: Mass = 0.3333 × 44 ≈ 14.67 g ≈ 14.7 g. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

In 2HNO₃ + 3H₂S -> 2NO + 3S + 4H₂O, what is the mass of S produced from 63 g of HNO₃ ? (Molar masses: HNO₃ = 63, S = 32)

Given: In 2HNO₃ + 3H₂S -> 2NO + 3S + 4H₂O, what is the mass of S produced from 63 g of HNO₃ ? (Molar masses: HNO₃ = 63, S = 32) Formula: Moles of HNO₃ = 63 / 63 = 1. Substitution & Calculation: 2 moles HNO₃ produce 3 moles S, so 1 mole produces 3/2 = 1.5 moles = 1.5 × 32 = 48 g. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

How many grams of O₂ are required to react with 14 g of N₂ to form N₂O₅? (Atomic masses: N = 14, O = 16)

Given: How many grams of O₂ are required to react with 14 g of N₂ to form N₂O₅? (Atomic masses: N = 14, O = 16) Formula: Moles of N₂ = 14 / 28 = 0.5 mol. Substitution & Calculation: Reaction: 2N₂ + 5O₂ -> 2N₂O₅ . . 2 mol N₂ require 5 mol O₂. Moles of O₂ = (5 / 2) × 0.5 = 1.25 mol. Mass = 1.25 × 32 = 40 g. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

In the reaction 4NH₃ + 5O₂ -> 4NO + 6H₂O, how many grams of O₂ are required to react with 17 g of NH₃? (Atomic

Given: In the reaction 4NH₃ + 5O₂ -> 4NO + 6H₂O, how many grams of O₂ are required to react with 17 g of NH₃? (Atomic masses: N = 14, H = 1, O = 16) These values define the system as per NCERT data. Formula: Molar mass of NH₃ = 17 g/mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles of NH₃ = 17 / 17 = 1 mol. 4 mol NH₃ require 5 mol O₂. Moles of O₂ = (5 / 4) × 1 = 1.25 mol. Mass = 1.25 × 32 = 40 g. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

What is the mass percentage of sulfur in Na₂S? (Atomic masses: Na = 23, S = 32)

Given: What is the mass percentage of sulfur in Na₂S? (Atomic masses: Na = 23, S = 32) Formula: Molar mass = (2 × 23) + 32 = 46 + 32 = 78 g/mol. Substitution & Calculation: Mass of S = 32 g. % S = (32 / 78) × 100 ≈ 41.03%. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

How many grams of NH₃ are produced from 14 g of N₂ with excess H₂ in the reaction N₂ + 3H₂ -> 2NH₃ ? (Atomic

Given: How many grams of NH₃ are produced from 14 g of N₂ with excess H₂ in the reaction N₂ + 3H₂ -> 2NH₃ ? (Atomic masses: N = 14, H = 1) These values define the system as per NCERT data. Formula: Moles of N₂ = 14 / 28 = 0.5 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 1 mol N₂ produces 2 mol NH₃. Moles of NH₃ = 0.5 × 2 = 1 mol. Molar mass of NH₃ = 17 g/mol. Mass = 1 × 17 = 17 g. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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