When 4.0 g of NH_3 reacts with 8.0 g of O_2 according to the equation 4 NH_3 + 5 O_2 → 4 NO + 6 H_2O, which substance is the limiting reactant?

• A. NO
• B. NH_3
• C. O_2
• D. H_2O

Answer: (B)

Verified step by step guidance

1. Write down the balanced chemical equation: 4 NH\_3 + 5 O\_2 \rightarrow 4 NO + 6 H\_2O.
2. Calculate the molar mass of NH\_3 and O\_2. For NH\_3, sum the atomic masses of 1 nitrogen and 3 hydrogens. For O\_2, multiply the atomic mass of oxygen by 2.
3. Convert the given masses of NH\_3 (4.0 g) and O\_2 (8.0 g) to moles using the formula: moles = mass / molar mass.
4. Use the mole ratio from the balanced equation to determine how many moles of O\_2 are required to react completely with the moles of NH\_3 calculated, and vice versa.
5. Compare the actual moles of each reactant to the required moles based on the stoichiometric ratio to identify which reactant is limiting (the one that will be completely consumed first).