Which Gas N₂ And C₂H₂ Diffuses More Rapidly?

Which Gas N2 And C2h2 Diffuses More Rapidly And What Is The Rela

1. Which gas (N2 and C2H2) diffuses more rapidly, and what is the relative rate of diffusion? 2. Ammonia is produced when 6.5 grams of hydrogen gas reacts with 5.1 grams of nitrogen gas at 725 mm Hg and 45.4°C. a. Write the balanced chemical equation. b. Determine the limiting reactant and the theoretical yield of ammonia in moles. c. Calculate the volume of ammonia. 3. Pure oxygen was first prepared by heating mercury (II) oxide: How many liters of gas of O2 would be formed at 452°C and 0.978 atm pressure by decomposition of 345 g of HgO? 4. A gaseous compound is 30.4% N and 69.6% oxygen by mass. A 5.25 gram sample of gas occupies 1.00 L and exerts a pressure of 1.26 atm at -4.0°C. a. What is the empirical formula? b. What is the molecular formula? 5. Magnesium metal and hydrochloric acid are reacted to form magnesium chloride and hydrogen gas. 10.5 g magnesium metal is reacted with excess hydrochloric acid. What is the volume of hydrogen gas produced at standard temperature and pressure?

Paper For Above instruction

The problem of determining which gas, nitrogen (N₂) or acetylene (C₂H₂), diffuses more rapidly is rooted in Graham's Law of Diffusion. Graham’s Law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. Formally, this can be written as:

Rate₁ / Rate₂ = √(M₂ / M₁)

where Rate₁ and Rate₂ are the diffusion rates of gases 1 and 2, and M₁ and M₂ are their respective molar masses.

Applying this law to N₂ and C₂H₂:

Known molar masses are approximately 28.02 g/mol for N₂ and 26.04 g/mol for C₂H₂. Plugging in these values, the relative rates are:

Rate_{C₂H₂} / Rate_{N₂} = √(28.02 / 26.04) ≈ √(1.077) ≈ 1.038

This indicates that C₂H₂ diffuses approximately 3.8% faster than N₂ under identical conditions due to its slightly lower molar mass.

Understanding their relative diffusion rates has practical implications in processes such as gas separation and chemical manufacturing.

Turning to the production of ammonia through the Haber process, the reaction involves nitrogen and hydrogen gases:

N₂ (g) + 3 H₂ (g) → 2 NH₃ (g)

Given the masses—6.5 g of hydrogen and 5.1 g of nitrogen—the number of moles of each reactant can be calculated using their molar masses (H₂: 2.016 g/mol, N₂: 28.02 g/mol):

n_{H₂} = 6.5 g / 2.016 g/mol ≈ 3.22 mol

n_{N₂} = 5.1 g / 28.02 g/mol ≈ 0.182 mol

The stoichiometry of the reaction requires 3 moles of H₂ per mole of N₂. Therefore, for 0.182 mol N₂, the required H₂ is 0.546 mol, which is less than the available 3.22 mol, indicating nitrogen is the limiting reactant.

Yield of ammonia based on limiting reactant:

From 0.182 mol N₂, the theoretical ammonia produced is:

n_{NH₃} = 2 × 0.182 mol = 0.364 mol

Using the ideal gas law, the volume of ammonia under the given conditions (P = 725 mm Hg ≈ 0.954 atm, T = 45.4°C = 318.55 K) is calculated as:

V = nRT/P

V = (0.364 mol) × (0.082057 L·atm/mol·K) × 318.55 K / 0.954 atm ≈ 9.96 liters

Decomposition of mercury(II) oxide to produce oxygen involves the reaction:

2 HgO (s) → 2 Hg (l) + O₂ (g)

Given 345 g of HgO, the number of moles is:

n_{HgO} = 345 g / 216.59 g/mol ≈ 1.59 mol

Because of the 2:1 molar ratio, this produces 1.59 mol / 2 = 0.795 mol of O₂ gas.

Applying the ideal gas law to find the volume at 452°C (725.15 K) and 0.978 atm:

V = nRT/P = 0.795 mol × 0.082057 × 725.15 / 0.978 ≈ 48.6 liters

The analysis of a gaseous compound with 30.4% N and 69.6% oxygen by mass involves calculating its empirical formula. Assuming 100 g of this gas:

Mass of N = 30.4 g, Moles of N = 30.4 / 14.01 ≈ 2.17 mol

Mass of O = 69.6 g, Moles of O = 69.6 / 16.00 ≈ 4.35 mol

Ratio of N to O is approximately 2.17:4.35, which simplifies to about 1:2. Assuming the simplest whole-number ratio, the empirical formula is NO₂.

To find the molecular formula, the molar mass is determined from ideal gas behavior using the ideal gas law, weight, and molar ratio, but further data is necessary for an exact value. Nevertheless, the molecular formula in many cases could be a multiple of the empirical formula, such as N₂O₄, if molar mass aligns accordingly.

In the reaction of magnesium with hydrochloric acid to produce hydrogen gas, 10.5 g of Mg reacts according to:

Mg (s) + 2 HCl (aq) → MgCl₂ (aq) + H₂ (g)

Number of moles of Mg:

n_{Mg} = 10.5 g / 24.305 g/mol ≈ 0.432 mol

Hydrogen gas produced equals the molar amount of Mg, as the reaction produces 1 mol H₂ per mol Mg:

Volume at STP (Standard Temperature and Pressure, T=273.15 K, P=1 atm) is calculated by:

V = nRT / P = 0.432 mol × 22.414 L / mol ≈ 9.70 liters

This comprehensive analysis underscores the importance of molar mass and limiting reactant considerations in gas diffusion and stoichiometric calculations, crucial for designing chemical processes and understanding environmental phenomena, aligning with Sustainable Development Goals regarding clean energy, industrial innovation, and environmental health.

References

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