If 34.2 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 299 Kelvin and 1.21 atmospheres? Show all of the work used to solve this problem.

2 Li (s) + 2 H2O (l) yields 2 LiOH (aq) + H2 (g)

mols Li = grams/atomic mass

Using the coefficients in the balanced equation, convert mols Li to mols H2.
Use PV = nRT, substitute the conditions and mols and solve fr V in liters.

To solve this problem, we need to use stoichiometry, a method that relates the amounts of substances in a chemical reaction. Here are the step-by-step calculations:

1. Convert the mass of lithium (34.2 grams) to moles. To do this, we need to know the molar mass of lithium, which is 6.94 g/mol. Divide the mass by the molar mass:

34.2 g / 6.94 g/mol = 4.92 mol of Li

2. According to the balanced chemical equation, 2 moles of lithium produce 1 mole of hydrogen gas. So, we need to convert the moles of lithium to moles of hydrogen gas:

4.92 mol Li x (1 mol H2 / 2 mol Li) = 2.46 mol H2

3. Now, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

We can rearrange the formula to solve for volume (V):
V = nRT / P

Insert the values into the formula:
V = (2.46 mol)(0.0821 L·atm/mol·K)(299 K) / 1.21 atm

4. Calculate the volume of hydrogen gas in liters:

V = (2.46 mol)(0.0821 L·atm/mol·K)(299 K) / 1.21 atm ≈ 49.07 L

Therefore, approximately 49.07 liters of hydrogen gas can be produced when 34.2 grams of lithium reacts with excess water at 299 Kelvin and 1.21 atmospheres.