Question

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

Answer 1

Molecular weight of Lithium = 6.941 g/mol.

From the reaction you have 2 moles of Li for every mol of H2 --2:1

(24.6 g of Li)(1 mol of Li/6.941 g of Li) = 3.544 moles of Li involved

(3.544 moles Li)(1 moles H2/2 moles Li) = 1.772 moles of H2


Ideal Gas Law equation PV = nRT

Therefore to find the Volume
: n = 1.772 moles
P = 1.01 atm
T = 301 K
R = 0.08205746 (L atm)/(mol K). This is to keep units consistency.

Rearranging for Volume of H2

V = nRT/P --plug the values in
and?=Liters

Answer 2

where do i plug them in at?

Answer 3

@zpupster

Answer 4

to the eqution i gave you

V = nRT/P

Answer 5

so i take all of the numbers you gave me and plug them in and thats the answer @zpupster

Answer 6

yes

Answer 7

ok thankyou @zpupster

Answer 8

good job!!

Answer 9

what units go behind it
would it be latm/molK?
@zpupster

Answer 10

: n = 1.772 moles
P = 1.01 atm
T = 301 K
R = 0.08205746 (L atm)/(mol K). This is to keep units consistency.


you will end up with liters after cancelling

i have to go now take care!!

Answer 11

V = nRT / P
V = ( (1.772 mol) * ( 0.0821 L atm/mol K) * 301 K ) / 1.01 atm
V = 43.3 Liters