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?

Question

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)

Answer 1

This is a mole ratio problem - with the added twist of using the ideal gas law to convert the answer in moles to an answer in liters.

 

Convert the grams of Li to moles:

 

(34.2g)/(6.941 g/mol) = 4.927 moles of Li

 

Looking at the balanced eqn:

2 moles of Li react to give 1 mol of H2.  Using the ratio 2:1,

4.927 mol of Li will react to give 2.464 mol of H2.

 

Now use the ideal gas law, PV=nRT to solve for the volume the 2.464 mol of H2 will occupy:

 

PV=nRT

(1.21)(v) = (2.464)(0.0821)(299)

 

Solving for V algebraically, V=50.0L

 

Keep in mind, when using the ideal gas law you always have to check the units. The units you need are always dictated by the units of R. For R=0,0821, the units of R are liter-atm per mol-degree K. All volumes need to be in L, all pressures need to be in atm, all temps needs to be in degrees K.  If units are not these, you must convert before using the formula.