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Chemistry - Chemical Reactions and Stoichiometry (The Mole Concept)

Grade 10IB

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

🔑Concepts

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The mole (nn) is the SI unit for the amount of substance. One mole contains exactly 6.022×10236.022 \times 10^{23} elementary entities, a value known as Avogadro's constant (LL or NAN_A).

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Molar mass (MM) is the mass of one mole of a substance, expressed in g⋅mol−1g \cdot mol^{-1}. It is calculated by summing the relative atomic masses (ArA_r) of all atoms in a chemical formula.

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Stoichiometry uses the quantitative relationships between reactants and products in a balanced chemical equation. The coefficients in the equation represent the mole ratio of the substances involved.

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The Limiting Reactant is the reagent that is completely consumed first in a chemical reaction, limiting the amount of product that can be formed. Reactants in excess are those not fully consumed.

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Theoretical yield is the maximum amount of product that can be produced from a given amount of reactants, while the percentage yield is the ratio of the actual (experimental) yield to the theoretical yield: ActualTheoretical×100\frac{\text{Actual}}{\text{Theoretical}} \times 100.

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Molar concentration (cc) describes the amount of solute in a given volume of solution, typically expressed in mol⋅dm−3mol \cdot dm^{-3}.

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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. At STP (273 K,100 kPa273 \, K, 100 \, kPa), one mole of an ideal gas occupies 22.7 dm322.7 \, dm^3.

📐Formulae

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

n=NNAn = \frac{N}{N_A}

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

PV=nRTPV = nRT

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

Atom Economy=Molar Mass of Desired ProductTotal Molar Mass of All Reactants×100\text{Atom Economy} = \frac{\text{Molar Mass of Desired Product}}{\text{Total Molar Mass of All Reactants}} \times 100

💡Examples

Problem 1:

Calculate the mass of 0.250 mol0.250 \, mol of Carbon Dioxide (CO2CO_2).

Solution:

m=n×M=0.250 mol×(12.01+2×16.00) g⋅mol−1=0.250×44.01=11.0 gm = n \times M = 0.250 \, mol \times (12.01 + 2 \times 16.00) \, g \cdot mol^{-1} = 0.250 \times 44.01 = 11.0 \, g.

Explanation:

First, find the molar mass (MM) of CO2CO_2 by adding the atomic masses of Carbon and Oxygen. Then, use the formula m=n×Mm = n \times M.

Problem 2:

If 10.0 g10.0 \, g of Magnesium (MgMg) reacts with excess Hydrochloric Acid (HClHCl), what volume of Hydrogen gas (H2H_2) is produced at STP? Equation: Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g).

Solution:

n(Mg)=10.0 g24.31 g⋅mol−1≈0.411 moln(Mg) = \frac{10.0 \, g}{24.31 \, g \cdot mol^{-1}} \approx 0.411 \, mol. According to the ratio 1:11:1, n(H2)=0.411 moln(H_2) = 0.411 \, mol. V=n×Vm=0.411 mol×22.7 dm3⋅mol−1≈9.33 dm3V = n \times V_m = 0.411 \, mol \times 22.7 \, dm^3 \cdot mol^{-1} \approx 9.33 \, dm^3.

Explanation:

Convert the mass of Magnesium to moles. Use the stoichiometric ratio from the balanced equation (1 mole of MgMg produces 1 mole of H2H_2) to find the moles of H2H_2. Finally, multiply by the molar gas volume at STP.

Problem 3:

A solution is prepared by dissolving 5.85 g5.85 \, g of NaClNaCl in water to make 500 cm3500 \, cm^3 of solution. Calculate the concentration in mol⋅dm−3mol \cdot dm^{-3}.

Solution:

n(NaCl)=5.85 g58.44 g⋅mol−1=0.100 moln(NaCl) = \frac{5.85 \, g}{58.44 \, g \cdot mol^{-1}} = 0.100 \, mol. V=500 cm3=0.500 dm3V = 500 \, cm^3 = 0.500 \, dm^3. c=0.100 mol0.500 dm3=0.200 mol⋅dm−3c = \frac{0.100 \, mol}{0.500 \, dm^3} = 0.200 \, mol \cdot dm^{-3}.

Explanation:

First, find the number of moles of solute (NaClNaCl). Convert the volume from cm3cm^3 to dm3dm^3 by dividing by 1000. Use the concentration formula c=n/Vc = n/V.