If 24.6 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 301 Kelvin and 1.01 atmosph eres? Show all of the work used to solve this problem. 2Li (s) + 2H2O (l) yields 2LiOH (aq) + H2 (g)
1 answer:
Answer: 43.3 l Explanation: 1) Chemical equation: 2 Li(s) + 2 H₂O (l) → 2LiOH(aq) + H₂ (g)2) Mole ratios: 2 mol Li : 2 mol H₂O : 2 mol LiOH : 1 mol H₂3) Number of moles of Li that react n = mass in grams / atomic mass = 24.6g / 6.941 g/mol = 3.54 moles4) Yield Proportion: 2 mol Li / 1 mol H₂ = 3.54 mol Li/ x ⇒ x = 3.54 mol Li × 1 mol H / 2 mol Li = 1.77 mol H₂4) Ideal gas equation PV = nRT ⇒ V = nRT / PV = 1.77 mol × 0.0821 [atm×l / (mol×K)] × 301 K / 1.01 atm = 43.3 l V = 43.3 l ← answer
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