Question

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

@taramgrant0543664

Answer 2

So your first step is to determine the moles of lithium to do this you take the mass (21.3g) divided by the molar mass of lithium

Answer 3

so i take 21.3/ 6.94

Answer 4

Yep so you get 3 moles of lithium reacting with excess water

Answer 5

So now we look at the stoichimometric coefficients. The ratio between lithium to hydrogen gas is 2:1. We found in the previous step that there are 3 moles of lithium so using the ratio we can relate to determine how many moles of hydrogen gas are produced. So how many moles of hydrogen gas are produced

Answer 6

3 moles lithium x 1H / 2Li = 1.52moles H. Correct?

Answer 7

Yes! So now you can put it into the ideal gas formula PV=nRT where P=1.4atm, n is the moles you solved for, R is the ideal gas constant and T is your temperature and solve for V your volume

Answer 8

R would be 0.082 thought I should mention that too

Answer 9

ok so the formula would be v = nrt/p. plugging that in would be 1.52moles x 0.082 x 297K / 1.4atm

Answer 10

Yes

Answer 11

then which would = 20.97 right?

Answer 12

20.87*

Answer 13

wait I did something wrong i got 26.44

Answer 14

Ya I got 26.4 I was trying to figure out how you got that other number lol

Answer 15

im not too sure either haha but thank you!

Answer 16

Haha happens to everyone! Happy to help!!