Answer:
28.5 g NH₃
Step-by-step explanation:
You can determine how much ammonia will form by identifying which reactant is the limiting reagent. This can be done by converting each reactant into moles, using the mole-to-mole ratio (after balancing the reaction), then converting ammonia to grams.
2 NO + 3 H₂ --> 2 NH₃ + O₂
Molar Mass (NO) = 14.01 + 16.00 = 30.01 g/mol
Molar Mass (H₂) = 2(1.008) = 2.016 g/mol
Molar Mass (NH₃) = 14.01 + 3 (1.008) = 17.034 g/mol
50.2 g NO 1 mol 2 mol NH₃ 17.034 g
---------------- x --------------- x -------------------- x ---------------- = 28.5 g NH₃
30.01 g 2 mol NO 1 mol
15.1 g H₂ 1 mol 2 mol NH₃ 17.034 g
---------------- x --------------- x -------------------- x ---------------- = 85.1 g NH₃
2.016 g 3 mol H₂ 1 mol
Nitrogen oxide (NO) is the limiting reagent because it produces the smallest amount of product. In other words, it runs out before all of the hydrogen gas (H₂) can be used. This means that the largest amount of ammonia that can be produced is 28.5 grams.