How many grams of oxygen (O2) are needed to completely react with 1326.76 grams of ammonia (NH3) according to the balanced chemical equation: 4 NH3 + 5 O2 → 4 NO + 6 H2O?

• A. 1000 grams
• B. 2000 grams
• C. 1568 grams
• D. 3200 grams

Answer: (C)

Verified step by step guidance

1. Identify the balanced chemical equation: 4 NH3 + 5 O2 → 4 NO + 6 H2O. This tells us that 4 moles of NH3 react with 5 moles of O2.
2. Calculate the molar mass of NH3. The molar mass of nitrogen (N) is approximately 14.01 g/mol and hydrogen (H) is approximately 1.01 g/mol. Therefore, the molar mass of NH3 is 14.01 + (3 * 1.01) g/mol.
3. Determine the number of moles of NH3 in 1326.76 grams. Use the formula: \( \text{moles of NH3} = \frac{\text{mass of NH3}}{\text{molar mass of NH3}} \).
4. Use the stoichiometry of the balanced equation to find the moles of O2 needed. According to the equation, 4 moles of NH3 react with 5 moles of O2. Therefore, \( \text{moles of O2} = \frac{5}{4} \times \text{moles of NH3} \).
5. Calculate the mass of O2 required using its molar mass. The molar mass of O2 is approximately 32.00 g/mol. Use the formula: \( \text{mass of O2} = \text{moles of O2} \times \text{molar mass of O2} \).