Answer to Question #146423 in General Chemistry for Lily

Question

Dinitrogen tetroxide decomposes to form nitrogen dioxide according to the following unbalanced equation: N2O4 (g) —-> NO2 (g)
If we start with 31.7g of dinitrogen tetroxide , what is the percent of yield of the reaction if we produce 10.2g of nitrogen dioxide?

Answer 1

Solution:

The balanced chemical equation:

N₂O₄(g) --> 2NO₂(g)

According to the chemical equation: n(N₂O₄) = n(NO₂)/2

Moles of dinitrogen tetroxide = n(N₂O₄) = Mass of N₂O₄ / Molar mass of N₂O₄

The molar mass of N₂O₄ is 92 g mol^-1.

n(N₂O₄) = (31.7 g ) / (92 g mol^-1) = 0.3446 mol

n(NO₂) = n(N₂O₄) × 2 = 0.3446 mol × 2 = 0.6892 mol

Moles of nitrogen dioxide = n(NO₂) = Mass of NO₂ / Molar mass of NO₂

The molar mass of NO₂ is 46 g mol^-1.

The theoretical mass of NO₂ = n(NO₂) × M(NO₂) = 0.6892 mol × 46 g mol^-1 = 31.7 g

The theoretical mass of NO₂ = 31.7 g

The actual mass of NO₂ = 10.2 g

% yield = [(actual mass) / (theoretical mass)] ×100%

% yield = (10.2 g / 31.7 g) × 100% = 32.1766% = 32.18%

% yield = 32.18%

OR

According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.

Hence,

Mass of NO₂ = Mass of N₂O₄

The theoretical mass of NO₂ = Mass of N₂O₄ = 31.7 g

The actual mass of NO₂ = 10.2 g

% yield = [(actual mass) / (theoretical mass)] ×100%

% yield = (10.2 g / 31.7 g) × 100% = 32.1766% = 32.18%

% yield = 32.18%

Answer: 32.18% is the percent of yield of the reaction.