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 atmospheres? show all of the work used to solve this problem. 2li (s) + 2h2o (l) yields 2lioh (aq) + h2 (g)

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 H₂

3) Number of moles of Li that react

n = mass in grams / atomic mass = 24.6g / 6.941 g/mol = 3.54 moles

4) 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 H₂

4) Ideal gas equation

PV = nRT ⇒ V = nRT / P

V = 1.77 mol × 0.0821 [atm×l / (mol×K)] × 301 K / 1.01 atm = 43.3 l

V = 43.3 l ← answer

You didn't post a list of words.