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Stoichiometric Relationships - Reacting masses and volumes

Grade 12IBChemistry

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

🔑Concepts

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The Law of Conservation of Mass states that in a chemical reaction, the mass of the products is equal to the mass of the reactants.

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Stoichiometry involves using the balanced chemical equation to determine the molar ratios between reactants and products.

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The limiting reactant is the substance that is entirely consumed in a reaction and determines the theoretical yield of products.

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The theoretical yield is the maximum amount of product that can be formed, whereas the actual yield is the amount actually obtained experimentally.

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Percentage yield is calculated as Actual YieldTheoretical Yield×100%\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%.

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Avogadro's Law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules.

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Molar volume (VmV_m) of an ideal gas at STP (273K273 K, 100kPa100 kPa) is 22.7dm3mol−122.7 dm^3 mol^{-1}.

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The Ideal Gas Law equation is PV=nRTPV = nRT, where pressure PP must be in PaPa, volume VV in m3m^3, and temperature TT in KK.

📐Formulae

n=mMn = \frac{m}{M}

n=VVmn = \frac{V}{V_m}

PV=nRTPV = nRT

c=nVc = \frac{n}{V}

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

P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}

💡Examples

Problem 1:

Calculate the volume of CO2CO_2 gas produced at STP when 5.00g5.00 g of CaCO3CaCO_3 reacts completely with excess HClHCl. The equation is: CaCO3(s)+2HCl(aq)→CaCl2(aq)+H2O(l)+CO2(g)CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g).

Solution:

n(CaCO3)=5.00g100.09gmol−1=0.04995moln(CaCO_3) = \frac{5.00 g}{100.09 g mol^{-1}} = 0.04995 mol. From the equation, the ratio of CaCO3CaCO_3 to CO2CO_2 is 1:11:1, so n(CO2)=0.04995moln(CO_2) = 0.04995 mol. Volume V=n×Vm=0.04995mol×22.7dm3mol−1=1.13dm3V = n \times V_m = 0.04995 mol \times 22.7 dm^3 mol^{-1} = 1.13 dm^3.

Explanation:

First, find the moles of the known substance (CaCO3CaCO_3). Use the stoichiometric ratio from the balanced equation to find the moles of CO2CO_2, then multiply by the molar volume at STP (22.7dm3mol−122.7 dm^3 mol^{-1}).

Problem 2:

Identify the limiting reactant when 2.0mol2.0 mol of H2H_2 reacts with 1.5mol1.5 mol of O2O_2 to form H2OH_2O.

Solution:

The balanced equation is 2H2(g)+O2(g)→2H2O(l)2H_2(g) + O_2(g) \rightarrow 2H_2O(l). Moles of H2H_2 required for 1.5mol1.5 mol of O2O_2 is 1.5×2=3.0mol1.5 \times 2 = 3.0 mol. Since we only have 2.0mol2.0 mol of H2H_2, H2H_2 is the limiting reactant.

Explanation:

Compare the available moles to the required moles based on the stoichiometric coefficients. The reactant that yields the least amount of product is the limiting reactant.

Problem 3:

A gas occupies 2.00dm32.00 dm^3 at 298K298 K and 100kPa100 kPa. Calculate its volume at 350K350 K and 150kPa150 kPa.

Solution:

Using P1V1T1=P2V2T2\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}: V2=P1V1T2T1P2=100kPa×2.00dm3×350K298K×150kPa=1.57dm3V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{100 kPa \times 2.00 dm^3 \times 350 K}{298 K \times 150 kPa} = 1.57 dm^3.

Explanation:

Apply the combined gas law, ensuring all units are consistent and temperature is in Kelvin (KK).

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