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Some Basic Concepts of Chemistry - Stoichiometry and Stoichiometric Calculations

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Stoichiometry is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), dealing with the calculation of masses and volumes of reactants and products in a chemical reaction.

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A balanced chemical equation is the foundation of stoichiometric calculations, adhering to the Law of Conservation of Mass.

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The coefficients in a balanced equation are called stoichiometric coefficients. They represent the number of moles (or molecules) participating in the reaction.

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Stoichiometric relationships can be categorized into: Mass-Mass, Mass-Volume, and Volume-Volume relationships.

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The Limiting Reagent is the reactant that is completely consumed first in a chemical reaction, thereby limiting the amount of product formed.

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The Excess Reagent is the reactant that remains unreacted after the limiting reagent is completely exhausted.

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Theoretical Yield is the maximum amount of product that can be formed from given amounts of reactants, while the Actual Yield is the amount of product actually obtained in the laboratory.

📐Formulae

Moles (n)=Mass (m)Molar Mass (M)\text{Moles (n)} = \frac{\text{Mass (m)}}{\text{Molar Mass (M)}}

Number of particles=n×NA, where NA=6.022×1023\text{Number of particles} = n \times N_A \text{, where } N_A = 6.022 \times 10^{23}

Volume of gas at STP (273.15 K, 1 bar)=n×22.7 L/mol\text{Volume of gas at STP (273.15 K, 1 bar)} = n \times 22.7 \text{ L/mol}

Stoichiometric Ratio=Moles of Substance AStoichiometric Coefficient of A\text{Stoichiometric Ratio} = \frac{\text{Moles of Substance A}}{\text{Stoichiometric Coefficient of A}}

% Yield=(Actual YieldTheoretical Yield)×100\% \text{ Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100

💡Examples

Problem 1:

Calculate the mass of CO2\text{CO}_2 produced by the complete combustion of 16 g16 \text{ g} of methane (CH4\text{CH}_4).

Solution:

  1. Write the balanced equation: CH4(g)+2O2(g)→CO2(g)+2H2O(g)\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g).
  2. Moles of CH4=16 g16 g/mol=1 mol\text{CH}_4 = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol}.
  3. From the stoichiometry, 1 mol1 \text{ mol} of CH4\text{CH}_4 produces 1 mol1 \text{ mol} of CO2\text{CO}_2.
  4. Mass of CO2=1 mol×44 g/mol=44 g\text{CO}_2 = 1 \text{ mol} \times 44 \text{ g/mol} = 44 \text{ g}.

Explanation:

Using the mass-mass relationship, we convert the given mass of the reactant to moles, use the stoichiometric ratio from the balanced equation to find moles of the product, and then convert those moles back to mass.

Problem 2:

50.0 kg50.0 \text{ kg} of N2(g)\text{N}_2 (g) and 10.0 kg10.0 \text{ kg} of H2(g)\text{H}_2 (g) are mixed to produce NH3(g)\text{NH}_3 (g). Identify the limiting reagent and calculate the amount of NH3\text{NH}_3 formed.

Solution:

  1. Balanced Equation: N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g).
  2. Moles of N2=50000 g28.0 g/mol≈1785.7 mol\text{N}_2 = \frac{50000 \text{ g}}{28.0 \text{ g/mol}} \approx 1785.7 \text{ mol}.
  3. Moles of H2=10000 g2.016 g/mol≈4960.3 mol\text{H}_2 = \frac{10000 \text{ g}}{2.016 \text{ g/mol}} \approx 4960.3 \text{ mol}.
  4. Required moles of H2\text{H}_2 for 1785.7 mol1785.7 \text{ mol} of N2=1785.7×3=5357.1 mol\text{N}_2 = 1785.7 \times 3 = 5357.1 \text{ mol}.
  5. Since we only have 4960.3 mol4960.3 \text{ mol} of H2\text{H}_2, H2\text{H}_2 is the limiting reagent.
  6. Moles of NH3\text{NH}_3 produced =23×4960.3≈3306.9 mol= \frac{2}{3} \times 4960.3 \approx 3306.9 \text{ mol}.
  7. Mass of NH3=3306.9 mol×17.03 g/mol≈56.3 kg\text{NH}_3 = 3306.9 \text{ mol} \times 17.03 \text{ g/mol} \approx 56.3 \text{ kg}.

Explanation:

The limiting reagent is identified by comparing the available moles to the required stoichiometric ratio. All product calculations are then based strictly on the quantity of the limiting reagent.

Stoichiometry and Stoichiometric Calculations Revision - Class 11 Chemistry CBSE