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Nitrogen dioxide reacts with water to produce oxygen and ammonia according to the following equation: 4NO₂(g) + 6H₂O(g) → 7O₂(g) + 4NH₃(g). How many grams of NH3 can be produced when 4.50 L of NO2 reacts at 405°C and 735 mmHg?

a) 19.08 g
b) 31.80 g
c) 12.72 g
d) 27.55 g

User Red
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Final answer:

The question asks to find the mass of ammonia produced from a reaction of nitrogen dioxide and water, using stoichiometry and the ideal gas law for calculations.

Step-by-step explanation:

The student's question involves a chemical reaction where nitrogen dioxide reacts with water to produce oxygen and ammonia. We're given the balanced equation 4NO₂(g) + 6H₂O(g) → 7O₂(g) + 4NH₃(g) and asked to calculate the mass of NH₃ that can be produced from 4.50 L of NO₂ at 405°C and 735 mmHg.

First, we convert the given volume of NO₂ into moles using the ideal gas law PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature in Kelvin. We then use the balanced equation to find the molar ratio between NO₂ and NH₃ and calculate the moles of NH₃ that will be produced. Finally, we convert moles of NH₃ into grams using the molar mass of NH₃.

Without doing the actual calculation, the process would involve using the appropriate R value for mmHg and liters, converting the temperature to Kelvin, and performing stoichiometric calculations to find the grams of NH₃ produced. The required R constant value would be 62.36 L mmHg mol⁻¹ K⁻¹, and the molar mass of NH₃ is 17.03 g/mol.

User Zhao Li
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