If 34.2 grams of lithium react with excess water, how many liters of hydrogen gas can be produced at 299 Kelvin and 1.21 atmospheres? Show all of the work used to solve this problem.
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someone? pleaz? I'm about to take a quiz and I don't understand.
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sorry idk :(
thx for looking at it :)
np :) wish i could help though..
balanced reaction first
sorry don't now super sorry their is a medal thoe
balanced reaction...like when you have those two group of numbers on either side of the equal sign and you balance them?
it shouldn't be an equals sign, but you're basically on the right idea. You need to turn that word problem into a chemical equation with the proper formulas, and then balance
how do I do that?
if you're doing this kind of problem and you don't know how to balance a reaction, you've got a lot of work to do
"lithium reacts with water" turn those words into proper chemical symbols
Li + H2O?
good, those are the reactants
the products are actually not all given to you. hydrogen gas will be formed, which is \(H_2\), but the other product is missing. This is a double displacement reaction
do you know what the other product is going to be?
the other product would by Lithium and Oxygen right? Lithium oxide?
almost, it's really lithium hydroxide, \(LiOH\)
not put all 4 of those together into a reaction
Li + H2O --- H2 + LiOH
good, now balance
Li2 + 2 H2O = 2 LiOH + H2
oops, i didn't mean to use =
now I have to find out how many moles I have right?
right
so I got 2.48 moles of Lithium, for the H2O do I have 2.48 moles or do I have to times that by 2 because of the 2 before it?
you don't need to know about the water, it's asking about the hydrogen \(produced\) from the lithium
oh ok
2.48 moles of lithium is not correct. How did you get it?
\[34.2\cancel{g \space Li} * (\frac{1 mol \space Li}{6.97\cancel{g \space Li}})\]
then I bet you multiplied by 2 because of the coefficient in the reaction?
the atomic mass is 6.9 when rounded. Since there's a 2 next to Li I timesed it by 2 and got 13.8. It said there was 34.2 grams of lithium so I divided 34.2 by 13.8. I ended up with2.48
yeah, don't do that
ok
the coefficients are handled later, molar mass is always grams per \(one\) mole, regardless of the coefficient in front of that particular reaction
oh ok
the proper conversion is the one I posted above
so it would just be 6.97?
no, it's \[34.2\cancel{g \space Li} * (\frac{1 mol \space Li}{6.97\cancel{g \space Li}})\]
4.9
yes, moles of Li
\(NOW\) the coefficients matter
your balanced reaction should be\[2Li + 2H_2O \rightarrow H_2 + 2LiOH\]
ok
the mole ratio is now (properly) 2:1, so that becomes the new conversion\[4.96\cancel{moles \space Li} * (\frac{1mole \space H_2}{2\cancel{moles \space Li}})\]
so 9.92?
you multiplied
you need to divide
divide 1 by 2?
divide the 4.96 by 2
oh, I'm sorry that looked like it was multiplying
2.48
2.48 moles of \(H_2\)
Ok, so now I know how many moles I have of Lithium and H2, now what do I do with them?
the moles of lithium are gone, used up
that's why the reaction arrow isn't an equals sign
what you have now is moles of \(H_2\)
but the problem is asking for \(LITERS\) of \(H_2\), which is a volume of gas. You need to use the ideal gas law
ohh, ok...pV=nRT?
yes. You're given a pressure, a temperature, and now you've just solved for the moles. R is a constant, so just find the volume
rearrange the equation first, and that will tell you which terms should be multiplied and which should be divided
ok
\[V = nRT \div P\]
true. now just plug in everything you have
ok, I got 50.31 as my answer
that's what I get, GJ
Thank you so much, now I think I know how to do that, and sorry for taking so long, I'm usually really smart but chemistry is my worst subject
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