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If 15 grams of iron react thus: Fe + H2SO4 = FeSO4 + H2, how many liters of hydrogen are liberated at normal conditions?

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Explanation

This problem consists of three parts:

  • First, determining the volume of hydrogen gas evolved from a known mass of iron in a reaction.
  • Second, converting the composition of a gas mixture (from brine electrolysis) from weight percentages to volume percentages using the ideal gas law.
  • Third, calculating the density of this gas mixture under specific conditions.

Question 1

Concepts

stoichiometry, mole concept, molar volume of gases (at STP), chemical equations

Explanation

The reaction: Fe+H2​SO4​→FeSO4​+H2​ shows that 1 mole of iron produces 1 mole of hydrogen gas. Using the molar mass of iron, we can find moles of Fe and thus moles (and volume) of H₂ evolved.

Step-By-Step Solution

Step 1: Calculate moles of Fe

Given, mass of Fe = 15 g. Molar mass of Fe = 55.85 g/mol.

Number of moles of Fe = 55.8515​=0.2686 mol (rounded to 4 decimal places).

Step 2: Moles of H₂ liberated

According to the balanced equation, 1 mol Fe → 1 mol H₂. So, moles of H₂ = 0.2686 mol.

Step 3: Convert moles of H₂ to volume (at NTP)

At NTP (Normal conditions: 0°C, 1 atm), 1 mol of gas = 22.4 L.

Volume of H₂ = moles × molar volume = 0.2686 × 22.4 = 6.016 L.

Final Answer

6.02 liters (rounded to 2 decimal places) of H2​ are liberated at normal conditions.

Question 2

Concepts

ideal gas law, Dalton’s Law of Partial Pressures, conversion from mass percent to mole (volume) percent, molar mass

Explanation

To convert mass % to volume (mole) %, use:

  • For each component, calculate moles from given mass, using molar mass.
  • The gas volume % is proportional to mole % for ideal gases.

Given composition by weight:

  • Cl₂: 67%, Br₂: 28%, O₂: 5% (assume 100 g total for convenience)

Step-By-Step Solution

Step 1: Calculate moles of each component

Molar masses:

  • Cl₂: 70.90 g/mol
  • Br₂: 159.8 g/mol
  • O₂: 32.00 g/mol

Moles of Cl₂ = 67 / 70.90 = 0.945 mol Moles of Br₂ = 28 / 159.8 = 0.175 mol Moles of O₂ = 5 / 32.00 = 0.156 mol

Step 2: Total moles

Total moles = 0.945 + 0.175 + 0.156 = 1.276 mol

Step 3: Volume (mole) % of each gas

  • Cl₂: (0.945/1.276) × 100% = 74.1%
  • Br₂: (0.175/1.276) × 100% = 13.7%
  • O₂: (0.156/1.276) × 100% = 12.2%

Final Answer

Composition by volume:

  • Chlorine (Cl₂): 74.1%
  • Bromine (Br₂): 13.7%
  • Oxygen (O₂): 12.2%

Question 3

Concepts

ideal gas law, density of gases, partial pressure, molar mass, Dalton’s law

Explanation

Density (d) can be calculated using the formula: d=RTPM​ where

  • P = Pressure (in atm),
  • M = average molar mass of mixture (g/mol)
  • R = 0.0821 L·atm·mol⁻¹·K⁻¹
  • T = Temperature (in Kelvin)

Step-By-Step Solution

Step 1: Calculate average molar mass (M) of the gas mixture

Mole fractions:

  • Cl₂: 0.945/1.276 = 0.741
  • Br₂: 0.175/1.276 = 0.137
  • O₂: 0.156/1.276 = 0.122

Average M=(0.741×70.90)+(0.137×159.8)+(0.122×32.00) =52.55+21.89+3.90=78.34g/mol

Step 2: Convert pressure and temperature to correct units

  • Pressure, P=740 mm Hg =760740​ atm = 0.974 atm
  • Temperature, T=25∘C=273.15+25=298.15 K

Step 3: Apply ideal gas law for density

d=RTPM​=0.0821×298.150.974×78.34​

Calculate denominator:

  • 0.0821×298.15=24.482

Numerator:

  • 0.974×78.34=76.30

d=24.48276.30​=3.12g/L

Final Answer

Density of the mixture at 25°C and 740 mm Hg: 3.12g/L​