Determine the mass (in grams) of water formed when 15.3 L NH₃ (at 298 K and 1.50 atm) is reacted with 21.7 L of O₂ (at 323 K and 1.1 atm).
4 NH₃ (g) + 5 O₂ (g) → 4 NO (g) + 6 H₂O (g)

• A. 12 g
• B. 19 g
• C. 25 g
• D. 38 g
• E. 51 g

Answer: (B)

Verified step by step guidance

1. First, use the ideal gas law to calculate the number of moles of NH3. The ideal gas law is given by the equation: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is temperature in Kelvin.
2. Calculate the number of moles of NH3 using the given conditions: P = 1.50 atm, V = 15.3 L, and T = 298 K. Rearrange the ideal gas law to solve for n: n = PV / RT.
3. Next, calculate the number of moles of O2 using the ideal gas law with the given conditions: P = 1.1 atm, V = 21.7 L, and T = 323 K. Again, use the formula n = PV / RT.
4. Determine the limiting reactant by comparing the mole ratio of NH3 to O2 from the balanced chemical equation: 4 NH3 + 5 O2 → 4 NO + 6 H2O. Calculate the theoretical moles of water that can be formed from each reactant and identify the limiting reactant.
5. Finally, use the stoichiometry of the balanced equation to calculate the mass of water formed. From the limiting reactant, determine the moles of H2O produced and convert this to grams using the molar mass of water (approximately 18.02 g/mol).