krit.club logo

Stoichiometric Relationships - Reacting masses and volumes

Grade 11IBChemistry

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

🔑Concepts

•

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

•

Molar mass (MM) is the mass of one mole of a substance, expressed in g mol−1g \text{ mol}^{-1}. It is numerically equal to the relative atomic/molecular mass (ArA_r or MrM_r).

•

The limiting reactant is the reagent that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product formed (theoretical yield).

•

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

•

Standard Temperature and Pressure (STP) for IB Chemistry is 273 K273 \text{ K} and 100 kPa100 \text{ kPa}. At STP, the molar volume (VmV_m) of an ideal gas is 22.7 dm3 mol−122.7 \text{ dm}^3 \text{ mol}^{-1}.

•

The Ideal Gas Law, PV=nRTPV = nRT, describes the relationship between pressure, volume, temperature, and moles for an ideal gas, where R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}.

•

Molar concentration (cc) is the amount of solute (in moles) per unit volume of solution (in dm3\text{dm}^3). The unit is mol dm−3\text{mol dm}^{-3}.

📐Formulae

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

n=NLn = \frac{N}{L}

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\%

Percentage Atom Economy=Molar mass of desired productSum of molar masses of all reactants×100%\text{Percentage Atom Economy} = \frac{\text{Molar mass of desired product}}{\text{Sum of molar masses of all reactants}} \times 100\%

💡Examples

Problem 1:

Calculate the mass of CO2CO_2 produced when 10.0 g10.0 \text{ g} of CaCO3CaCO_3 is heated and undergoes thermal decomposition according to the equation: CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \rightarrow CaO(s) + CO_2(g).

Solution:

  1. Calculate moles of CaCO3CaCO_3: n(CaCO3)=mM=10.0 g100.09 g mol−1≈0.0999 moln(CaCO_3) = \frac{m}{M} = \frac{10.0 \text{ g}}{100.09 \text{ g mol}^{-1}} \approx 0.0999 \text{ mol}. \ 2. Use stoichiometric ratio (1:11:1): n(CO2)=n(CaCO3)=0.0999 moln(CO_2) = n(CaCO_3) = 0.0999 \text{ mol}. \ 3. Convert moles to mass: m(CO2)=n×M=0.0999 mol×44.01 g mol−1≈4.40 gm(CO_2) = n \times M = 0.0999 \text{ mol} \times 44.01 \text{ g mol}^{-1} \approx 4.40 \text{ g}.

Explanation:

First, find the number of moles of the given substance using its molar mass. Use the coefficients from the balanced equation to find the moles of the target substance. Finally, convert those moles back into mass.

Problem 2:

What volume of hydrogen gas, H2H_2, is produced at STP when 0.50 mol0.50 \text{ mol} of magnesium reacts with excess hydrochloric acid? Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g)Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)

Solution:

  1. Identify the molar ratio of MgMg to H2H_2: 1:11:1. \ 2. Therefore, n(H2)=0.50 moln(H_2) = 0.50 \text{ mol}. \ 3. Calculate volume at STP: V=n×Vm=0.50 mol×22.7 dm3 mol−1=11.35 dm3V = n \times V_m = 0.50 \text{ mol} \times 22.7 \text{ dm}^3 \text{ mol}^{-1} = 11.35 \text{ dm}^3.

Explanation:

Since the ratio of MgMg to H2H_2 is 1:11:1, 0.500.50 moles of MgMg produces 0.500.50 moles of H2H_2. At STP, one mole of any gas occupies 22.7 dm322.7 \text{ dm}^3.

Problem 3:

A gas occupies 2.00 dm32.00 \text{ dm}^3 at 298 K298 \text{ K} and 100 kPa100 \text{ kPa}. What will its volume be at 350 K350 \text{ K} if the pressure is increased to 150 kPa150 \text{ kPa}?

Solution:

Using the combined gas law: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}. \ V2=P1V1T2P2T1=100 kPa×2.00 dm3×350 K150 kPa×298 K≈1.57 dm3V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac{100 \text{ kPa} \times 2.00 \text{ dm}^3 \times 350 \text{ K}}{150 \text{ kPa} \times 298 \text{ K}} \approx 1.57 \text{ dm}^3.

Explanation:

The combined gas law allows for the calculation of changes in pressure, volume, or temperature when the amount of gas remains constant.

Reacting masses and volumes - Revision Notes & Key Formulas | IB Grade 11 Chemistry