Question

f 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) 2 LiOH (aq) + H2 (g)

Answer 1

Take note of the information we have: We have a pressure p, temperature T, and the information to find the number of moles.... sounds a lot like the ideal gas equation... lets write it up:
\[pV=nRT\]
R is the gas constant, n is the amount of moles, T the temperature, V the volume and p the pressure.
We want the volume so we isolate V in the equation:
\[V=\frac{ nRT }{ p }\]
Now what we don't now so far is the n... but we can calculate that:
"34.2 grams of lithium react with excess water" we write the equation:
2 Li(s) + 2 H2O -> 2 LiOH (aq) + H2(g)

Do the rest of the calculations between mass and the molecular mass to find the amount of molecules in moles. Take account for the stoichiometry and then put it into the ideal gas equation and you have the answer

Answer 2

As side note it is wise to say the answer in inaccurate due to the definitions of a ideal/perfect gas (remember it is a imaginary due to the fact it does not have any volume, but is considered point masses)

Answer 3

If trouble with the calculations let me know.

Answer 4

@Frostbite I'm having trouble solving it. Only the atm cancels out.

Answer 5

Okay I try do the calculations:

34.2 grams of lithium react with excess water:
2 Li (s) + 2 H2O (l) 2 LiOH (aq) + H2 (g)
\[\Large n(H _{2})=\frac{ n(Li) }{ 2 }\]
\[\Large n(Li)=\frac{ m(Li) }{ M(Li) }=\frac{ 34.2 g }{ 6.94 \frac{ g }{ mol } }=4.927 mol\]
\[\Large n(H _{2})=\frac{ 4.927 mol }{ 2 }=2.46 mol\]

We assume that molecular hydrogen act as a ideal gas and thereby write the ideal gas equation
\[\Large pV=nRT \to V=\frac{ nRT }{ P }\]
Insert the information.
\[\Large V=\frac{ 2.26 mol \times R \times 299 K }{ 1.21 atm }\]
The gas constant can take a lot of values depending on the units, the most impotent you are going to need is the relation to pressure in atm, pascal and bar.
\[\Large R=k \times N _{a}=8.31 \frac{ J }{ mol \times K } \approx 0.08205 \frac{ L \times atm }{ mol \times K }\]
\[\Large 1 atm \approx 101 kPa\]
So the volume most be:
\[\Large V=\frac{ 2.26 mol \times 0.08205 \frac{ L \times atm }{ mol \times K } \times 299 K }{ 1.21 atm }\]
Evaluate it (now a calculator would come in handy):
http://bit.ly/1cyDClo