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Stoichiometry - Reacting masses and volumes

Grade 12IGCSEChemistry

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

🔑Concepts

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The mole (nn) is the unit for the amount of substance. One mole contains Avogadro's constant (6.02×10236.02 \times 10^{23}) of particles.

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The Molar Mass (MrM_r) is the mass of one mole of a substance, calculated by summing the relative atomic masses (ArA_r) of all atoms in the chemical formula.

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Stoichiometry involves using the coefficients of a balanced chemical equation to find the mole ratio between reactants and products.

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Molar Gas Volume: At Room Temperature and Pressure (RTP), one mole of any gas occupies a volume of 24 dm324 \text{ dm}^3 (or 24,000 cm324,000 \text{ cm}^3).

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Concentration of a solution (cc) is defined as the amount of solute (moles) dissolved in a unit volume (VV) of solution, typically expressed in mol/dm3\text{mol/dm}^3.

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

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Percentage Yield measures the efficiency of a reaction: Percentage Yield=Actual YieldTheoretical Yield×100%\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%.

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Percentage Purity is the percentage of a specific substance in an impure sample: Percentage Purity=Mass of pure substanceTotal mass of sample×100%\text{Percentage Purity} = \frac{\text{Mass of pure substance}}{\text{Total mass of sample}} \times 100\%.

📐Formulae

n=mMrn = \frac{m}{M_r}

n=V24 (for gases at RTP, where V is in dm3)n = \frac{V}{24} \text{ (for gases at RTP, where } V \text{ is in dm}^3\text{)}

c=nV (where V is in dm3)c = \frac{n}{V} \text{ (where } V \text{ is in dm}^3\text{)}

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

Concentration (g/dm3)=Concentration (mol/dm3)×Mr\text{Concentration (g/dm}^3\text{)} = \text{Concentration (mol/dm}^3\text{)} \times M_r

💡Examples

Problem 1:

Calculate the mass of MgOMgO produced when 1.2 g1.2 \text{ g} of MgMg burns completely in oxygen.

Solution:

2Mg+O2→2MgO2Mg + O_2 \rightarrow 2MgO

  1. Find moles of MgMg: n(Mg)=1.224=0.05 moln(Mg) = \frac{1.2}{24} = 0.05 \text{ mol}.
  2. Use mole ratio (Mg:MgO=2:2=1:1Mg:MgO = 2:2 = 1:1): n(MgO)=0.05 moln(MgO) = 0.05 \text{ mol}.
  3. Calculate mass: m(MgO)=0.05×(24+16)=0.05×40=2.0 gm(MgO) = 0.05 \times (24 + 16) = 0.05 \times 40 = 2.0 \text{ g}.

Explanation:

First, the balanced equation is used to find the molar ratio. Then, the moles of the known substance are calculated to find the moles of the unknown substance.

Problem 2:

What volume of CO2CO_2 gas at RTP is produced from the thermal decomposition of 5.0 g5.0 \text{ g} of CaCO3CaCO_3?

Solution:

CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g)

  1. Mr(CaCO3)=40+12+(3×16)=100M_r(CaCO_3) = 40 + 12 + (3 \times 16) = 100.
  2. n(CaCO3)=5.0100=0.05 moln(CaCO_3) = \frac{5.0}{100} = 0.05 \text{ mol}.
  3. Ratio is 1:11:1, so n(CO2)=0.05 moln(CO_2) = 0.05 \text{ mol}.
  4. V(CO2)=0.05×24=1.2 dm3V(CO_2) = 0.05 \times 24 = 1.2 \text{ dm}^3.

Explanation:

Convert the mass of the solid reactant to moles, use the 1:11:1 stoichiometry to find moles of gas, and multiply by the molar gas volume constant (24 dm324 \text{ dm}^3).

Problem 3:

Calculate the concentration of a solution where 0.2 moles0.2 \text{ moles} of HClHCl are dissolved in 250 cm3250 \text{ cm}^3 of water.

Solution:

  1. Convert volume to dm3\text{dm}^3: V=2501000=0.25 dm3V = \frac{250}{1000} = 0.25 \text{ dm}^3.
  2. Apply formula: c=nV=0.20.25=0.8 mol/dm3c = \frac{n}{V} = \frac{0.2}{0.25} = 0.8 \text{ mol/dm}^3.

Explanation:

Always ensure the volume is converted from cm3\text{cm}^3 to dm3\text{dm}^3 by dividing by 10001000 before calculating concentration.

Reacting masses and volumes - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry