Question

Please Help?! I will fan/medal! Just check my work, please?!
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.

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

Answer 1

so I got:
2Li(s) + 2H2O(l) --> 2LiOH(aq) + H2(g)

21.3g Li /7g/mole = 3.04moles Li
3.04moles Li x (1H2 / 2Li) = 1.52moles H2

pv = nrt, solving for v we get
v = nrt/p
v = 1.52moles x 0.0821L-atm/mole-K x 297K / 1.4atm
v = 26.47L

Answer 2

is this correct?

Answer 3

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Answer 4

Your work looks correct.

Answer 5

thank you @abb0t
:D

Answer 6

looks right to me

Answer 7

Thank you :D