Lecture 5: Reacting Masses of Solids, Liquids, and Gases and Reacting Volumes of Gases

Reacting Masses and Gas Volumes

Reacting Masses

• Learning aims:
  • Calculate the mass of a product from the mass of a reactant (and vice versa).
  • Calculate gas volumes in reactions using the fact that one mole of gas is 24 dm^3 at room temperature and pressure.
• More details on the learning aims:
  • Understanding the stoichiometry of chemical reactions.
  • Applying the concept of molar mass to convert between mass and moles.
  • Using the ideal gas law to relate gas volumes to moles at room temperature and pressure.
• Key Formula:
  • Number of moles (n) = mass in grams (m) / molar mass in grams per mole (M)- n = \frac{m}{M}
  • Mass (m) = number of moles (n) * molar mass (M)- m = n \times M
  • Details on key formula:
  • The number of moles (n) is a fundamental unit in chemistry that relates the mass of a substance to its molar mass.
  • The molar mass (M) is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).
• Examples:
  • Iron Sulfide
  • Sodium Oxide
  • Oxygen from Hydrogen Peroxide
  • Fermentation

Example 1: Iron Sulfide

• Problem: Calculate the mass of iron sulfide (FeS) produced when 5.6 grams of iron (Fe) reacts completely with excess sulfur (S).
• Solution:
  1. Chemical equation: Fe + S \rightarrow FeS
  2. Calculate moles of iron:-
     n(Fe) = \frac{5.6 \, g}{56 \, g/mol} = 0.1 \, mol
  3. Ratio of Fe to FeS: 1:1
  4. Moles of iron sulfide: n(FeS) = 0.1 \, mol
  5. Convert moles of FeS to mass:-
     m(FeS) = 0.1 \, mol \times 88 \, g/mol = 8.8 \, g

Example 2: Sodium Oxide

• Calculate the mass of sodium oxide (Na_2O) produced from a given amount of reactants.

Example 3: Oxygen from Hydrogen Peroxide

• Calculate the mass of oxygen (O2) produced by decomposing hydrogen peroxide (H2O_2).

Example 4: Fermentation

• Write the equation for fermentation and perform calculations.

Gas Volumes and Moles

• Key relationship: At room temperature and pressure (RTP), one mole of any gas occupies 24 dm^3 (24,000 cm^3).
  • Standard conditions:
  • Standard Temperature and Pressure (STP) which is 0 degrees Celsius (273.15 K) and 1 atmosphere (101.325 kPa).
• Volume per mole of gas at RTP = 24 dm^3/mol
• Equations:
  • Number of moles of gas = Volume of gas in dm^3 / 24- n = \frac{V \,(dm^3)}{24}
  • Number of moles of gas = Volume of gas in cm^3 / 24,000- n = \frac{V \,(cm^3)}{24000}

Example 1: Moles of Oxygen Gas

• Problem: How many moles are there in 6 dm^3 of oxygen gas at RTP?
• Solution:-
  n(O_2) = \frac{6 \, dm^3}{24 \, dm^3/mol} = 0.25 \, mol

Finding Volume from Moles

• Rearranged formula: Volume = Number of moles * 24,000 (cm^3) or 24 (dm^3).-
  V = n \times 24000 \,(cm^3)

• V = n \times 24 \,(dm^3)

Example 2: Volume of a Gas

• Problem: What is the volume in cm^3 of 2.13 \times 10^{-3} moles of a gas at RTP?
• Solution:-
  V = 2.13 \times 10^{-3} \, mol \times 24000 \, cm^3/mol = 51.12 \, cm^3

Gas Volumes in Equations

• The volume of any gas is proportional to the number of moles of the gas.

Example: Ammonia Production

• Problem: What volume of ammonia (NH3) can be produced from 10 cm^3 of hydrogen (H2)?
• Solution:
  1. Chemical equation: N2 + 3H2 \rightarrow 2NH_3
  2. Ratio of H2 to NH3 is 3:2
  3. Volume of NH_3 produced: \frac{2}{3} \times 10 \, cm^3 = 6.7 \, cm^3
• Volume of nitrogen needed: \frac{1}{3} \times 10 \, cm^3 = 3.33 \, cm^3

Example: Sodium and Water

• Problem: Calculate the volume of hydrogen released in the reaction of 20 grams of sodium with water.
• Solution:
  1. Chemical equation: 2Na + 2H2O \rightarrow 2NaOH + H2
  2. Calculate moles of sodium:-
     $$n(Na) = \frac{20 \, g}{23 \, g/mol} = 0.87 \, mol