If 21.3 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 297 Kelvin and 1.40 atmospheres? Show all of the work used to solve this problem.

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

Question

Chemistry

Hannah

18 May, 06:48

If 21.3 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 297 Kelvin and 1.40 atmospheres? Show all of the work used to solve this problem.

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

+3

Answers (1)

Know the Answer?

Amaris Higgins · Answer 1

() = 2 (2)

() =

=

21.3 6.94

= 3.07

(2) = 1.53 (2) = (2) ∗

At STP: = 22.4 /:

(2,) = 1.53∗22.4 = 34.4

At 297 K and 1.4 atm:

= ⇒

1∗34.4 273.15

=

1.40∗ 297

= 26.7

which means the answer is 26.7 liters

If 21.3 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 297 Kelvin and 1.40 atmospheres? Show all of the work used to solve this problem.2Li (s) + 2H2O (l) - > 2LiOH (aq) + H2 (g)

If 21.3 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 297 Kelvin and 1.40 atmospheres? Show all of the work used to solve this problem.

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