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 ($O<em>2$) produced by decomposing hydrogen peroxide ($H</em>2O_2$).

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 ($NH<em>3$) can be produced from 10 $cm^3$ of hydrogen ($H</em>2$)?
• Solution:
  1. Chemical equation: $N<em>2 + 3H</em>2 \rightarrow 2NH_3$
  2. Ratio of $H<em>2$ to $NH</em>3$ 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: $2Na + 2H<em>2O \rightarrow 2NaOH + H</em>2$
  2. Calculate moles of sodium:-
     $$n(Na) = \frac{20 \, g}{23 \, g/mol} = 0.87 \, mol