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)

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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)

Chemistry
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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
so i take 21.3/ 6.94

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Yep so you get 3 moles of lithium reacting with excess water
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
3 moles lithium x 1H / 2Li = 1.52moles H. Correct?
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
R would be 0.082 thought I should mention that too
ok so the formula would be v = nrt/p. plugging that in would be 1.52moles x 0.082 x 297K / 1.4atm
Yes
then which would = 20.97 right?
20.87*
wait I did something wrong i got 26.44
Ya I got 26.4 I was trying to figure out how you got that other number lol
im not too sure either haha but thank you!
Haha happens to everyone! Happy to help!!

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