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Some Basic Concepts of Chemistry - Stoichiometry and Stoichiometric Calculations

Grade 11ICSEChemistry

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

🔑Concepts

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Stoichiometry is based on the Law of Conservation of Mass, where the total mass of reactants equals the total mass of products. It involves calculating the quantities of reactants and products in a balanced chemical reaction.

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Stoichiometric Coefficients: The numbers used to balance a chemical equation represent the molar ratios of the substances. For example, in N2(g)+3H2(g)ightarrow2NH3(g)N_2(g) + 3H_2(g) ightarrow 2NH_3(g), the coefficients are 1,3,1, 3, and 22.

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Mole-Mole Relationship: This is the primary bridge in stoichiometric calculations. If the moles of one substance are known, the moles of any other substance in the reaction can be found using the ratio of their coefficients.

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Mass-Mass Relationship: Involves converting the given mass of a substance to moles using n=mMn = \frac{m}{M}, using the molar ratio from the balanced equation, and then converting back to the mass of the desired product.

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Mass-Volume and Volume-Volume Relationship: For gaseous reactions at STP (Standard Temperature and Pressure), 11 mole of any gas occupies 22.4 L22.4 \text{ L}. Volume ratios are equivalent to mole ratios for gases at the same temperature and pressure.

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Limiting Reagent: The reactant that is completely consumed in a reaction is the limiting reagent. It limits the amount of product formed. Any reactant left over after the reaction is called the 'Excess Reagent'.

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Percentage Yield: In practice, the amount of product obtained (Actual Yield) is often less than the amount calculated (Theoretical Yield). It is calculated as: Percentage Yield=Actual YieldTheoretical Yield×100\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100.

📐Formulae

Number of moles (n)=Given mass in grams (m)Molar mass (M)\text{Number of moles } (n) = \frac{\text{Given mass in grams } (m)}{\text{Molar mass } (M)}

Volume of gas at STP (V)=n×22.4 L\text{Volume of gas at STP } (V) = n \times 22.4 \text{ L}

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

Actual Yield=Percentage Yield×Theoretical Yield100\text{Actual Yield} = \frac{\text{Percentage Yield} \times \text{Theoretical Yield}}{100}

\text{Mass % of an element} = \frac{\text{Mass of that element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100

💡Examples

Problem 1:

Calculate the mass of CO2CO_2 produced by the complete combustion of 16 g16 \text{ g} of methane (CH4CH_4).

Solution:

  1. Write the balanced equation: CH4(g)+2O2(g)→CO2(g)+2H2O(g)CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g).
  2. Moles of CH4=16 g16 g/mol=1 molCH_4 = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol}.
  3. From the equation, 11 mole of CH4CH_4 gives 11 mole of CO2CO_2.
  4. Mass of CO2=1 mol×44 g/mol=44 gCO_2 = 1 \text{ mol} \times 44 \text{ g/mol} = 44 \text{ g}.

Explanation:

We first convert the mass of the known reactant to moles, use the stoichiometric ratio (1:11:1) to find the moles of the product, and then convert those moles back into mass using the molar mass of CO2CO_2 (44 g/mol44 \text{ g/mol}).

Problem 2:

Identify the limiting reagent when 3 g3 \text{ g} of H2H_2 reacts with 29 g29 \text{ g} of O2O_2 to form H2OH_2O.

Solution:

  1. Balanced equation: 2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O.
  2. Moles of H2=32.016≈1.488 molH_2 = \frac{3}{2.016} \approx 1.488 \text{ mol}.
  3. Moles of O2=2932≈0.906 molO_2 = \frac{29}{32} \approx 0.906 \text{ mol}.
  4. Requirement: 22 moles of H2H_2 need 11 mole of O2O_2.
  5. 1.488 moles of H21.488 \text{ moles of } H_2 would require 1.4882=0.744 moles of O2\frac{1.488}{2} = 0.744 \text{ moles of } O_2.
  6. Since we have 0.906 mol0.906 \text{ mol} of O2O_2 (which is more than 0.7440.744), O2O_2 is in excess and H2H_2 is the limiting reagent.

Explanation:

The limiting reagent is found by comparing the actual mole ratio to the stoichiometric ratio. Since H2H_2 will be exhausted first based on the 2:12:1 requirement, it determines the extent of the reaction.

Stoichiometry and Stoichiometric Calculations Revision - Class 11 Chemistry ICSE