Given 3.00 g of N2H4 (MW 32.05) and 5.00 g of O2 (MW 32.00) react according to N2H4 + O2 -> N2 + 2 H2O. Which is the limiting reagent and how many grams of N2 are produced?

• A. Limiting reagent N2H4; N2 produced ≈ 2.62 g
• B. Limiting reagent O2; N2 produced ≈ 3.00 g
• C. Limiting reagent N2H4; N2 produced ≈ 4.00 g
• D. Limiting reagent none; N2 produced ≈ 1.00 g

Answer: (A)

The test is about identifying the limiting reagent and carrying out stoichiometric calculations. In this reaction, N2H4 and O2 react in a 1:1 mole ratio to produce N2 and H2O. Convert the given masses to moles: N2H4 is 3.00 g / 32.05 g/mol ≈ 0.0937 mol. O2 is 5.00 g / 32.00 g/mol = 0.15625 mol. Since the reaction needs equal moles of each reactant, N2H4 will run out first (0.0937 mol available vs 0.15625 mol required to consume all of O2). So N2H4 is the limiting reagent.

The amount of N2 produced is set by the amount of N2H4 that reacts: 0.0937 mol N2H4 → 0.0937 mol N2 (1:1). Convert to grams: 0.0937 mol × 28.02 g/mol ≈ 2.62 g N2.

Hence the limiting reagent is N2H4 and about 2.62 g of N2 are produced.