Question

ammonia gas reacts with oxygen gas, o2(g), to produce nitrogen dioxide and water. when 28.5 g of ammonia gas reacts with 83.4 g of oxygen gas to produce 61.9 g of nitrogen dioxide, what is the percent yield of the reaction?

Answer 1

The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.

A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)

NH₃ + O₂ → NO₂ + H₂O

The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O

Calculate the number of moles from the reactant

• Ammonia gas
  Molar mass N = 14 gr/mol
  Molar mass H = 1 gr/mol
  Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
  mass = 28.5 grams
  n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol
• Oxygen gas
  Molar mass O = 16 gr/mol
  Molar mass O₂ = 16 × 2 = 32 gr/mol
  mass = 83.4 grams
  n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol
• n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
  n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
  0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.

According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.

• Theoretically the number moles of NO₂
  n O₂ : n NO₂ = 7 : 4
  2.61 : n NO₂ = 7 : 4
  n NO₂ = 4 x 2.61 : 7 = 1.49 mol
• The actual number of moles NO₂
  Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
  n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol

The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.

P = (1.35 ÷ 1.49) × 100%

P = 0.9052 × 100%

P = 90.52%

Learn more about stoichiometry here: brainly.com/question/13691565

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