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Chemistry 19 Online
OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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OpenStudy (anonymous):

someone? pleaz? I'm about to take a quiz and I don't understand.

OpenStudy (anonymous):

@dumbsearch2 @DullJackel09 @Daniellelovee @DarkMoonZ

OpenStudy (anonymous):

@emo__gamer101 @elyrose101 @fluteprincess352 @Godlovesme @gelow @ghuczek @gahm8684 @hhelpplzzzz @HelpBlahBlahBlah @Hotchellerae21 @Here_to_Help15

OpenStudy (godlovesme):

sorry idk :(

OpenStudy (anonymous):

thx for looking at it :)

OpenStudy (godlovesme):

np :) wish i could help though..

OpenStudy (jfraser):

balanced reaction first

OpenStudy (gelow):

sorry don't now super sorry their is a medal thoe

OpenStudy (anonymous):

balanced reaction...like when you have those two group of numbers on either side of the equal sign and you balance them?

OpenStudy (jfraser):

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

OpenStudy (anonymous):

how do I do that?

OpenStudy (jfraser):

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

OpenStudy (jfraser):

"lithium reacts with water" turn those words into proper chemical symbols

OpenStudy (anonymous):

Li + H2O?

OpenStudy (jfraser):

good, those are the reactants

OpenStudy (jfraser):

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

OpenStudy (jfraser):

do you know what the other product is going to be?

OpenStudy (anonymous):

the other product would by Lithium and Oxygen right? Lithium oxide?

OpenStudy (jfraser):

almost, it's really lithium hydroxide, \(LiOH\)

OpenStudy (jfraser):

not put all 4 of those together into a reaction

OpenStudy (anonymous):

Li + H2O --- H2 + LiOH

OpenStudy (jfraser):

good, now balance

OpenStudy (anonymous):

Li2 + 2 H2O = 2 LiOH + H2

OpenStudy (anonymous):

oops, i didn't mean to use =

OpenStudy (anonymous):

now I have to find out how many moles I have right?

OpenStudy (jfraser):

right

OpenStudy (anonymous):

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?

OpenStudy (jfraser):

you don't need to know about the water, it's asking about the hydrogen \(produced\) from the lithium

OpenStudy (anonymous):

oh ok

OpenStudy (jfraser):

2.48 moles of lithium is not correct. How did you get it?

OpenStudy (jfraser):

\[34.2\cancel{g \space Li} * (\frac{1 mol \space Li}{6.97\cancel{g \space Li}})\]

OpenStudy (jfraser):

then I bet you multiplied by 2 because of the coefficient in the reaction?

OpenStudy (anonymous):

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

OpenStudy (jfraser):

yeah, don't do that

OpenStudy (anonymous):

ok

OpenStudy (jfraser):

the coefficients are handled later, molar mass is always grams per \(one\) mole, regardless of the coefficient in front of that particular reaction

OpenStudy (anonymous):

oh ok

OpenStudy (jfraser):

the proper conversion is the one I posted above

OpenStudy (anonymous):

so it would just be 6.97?

OpenStudy (jfraser):

no, it's \[34.2\cancel{g \space Li} * (\frac{1 mol \space Li}{6.97\cancel{g \space Li}})\]

OpenStudy (anonymous):

4.9

OpenStudy (jfraser):

yes, moles of Li

OpenStudy (jfraser):

\(NOW\) the coefficients matter

OpenStudy (jfraser):

your balanced reaction should be\[2Li + 2H_2O \rightarrow H_2 + 2LiOH\]

OpenStudy (anonymous):

ok

OpenStudy (jfraser):

the mole ratio is now (properly) 2:1, so that becomes the new conversion\[4.96\cancel{moles \space Li} * (\frac{1mole \space H_2}{2\cancel{moles \space Li}})\]

OpenStudy (anonymous):

so 9.92?

OpenStudy (jfraser):

you multiplied

OpenStudy (jfraser):

you need to divide

OpenStudy (anonymous):

divide 1 by 2?

OpenStudy (jfraser):

divide the 4.96 by 2

OpenStudy (anonymous):

oh, I'm sorry that looked like it was multiplying

OpenStudy (anonymous):

2.48

OpenStudy (jfraser):

2.48 moles of \(H_2\)

OpenStudy (anonymous):

Ok, so now I know how many moles I have of Lithium and H2, now what do I do with them?

OpenStudy (jfraser):

the moles of lithium are gone, used up

OpenStudy (jfraser):

that's why the reaction arrow isn't an equals sign

OpenStudy (jfraser):

what you have now is moles of \(H_2\)

OpenStudy (jfraser):

but the problem is asking for \(LITERS\) of \(H_2\), which is a volume of gas. You need to use the ideal gas law

OpenStudy (anonymous):

ohh, ok...pV=nRT?

OpenStudy (jfraser):

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

OpenStudy (jfraser):

rearrange the equation first, and that will tell you which terms should be multiplied and which should be divided

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

\[V = nRT \div P\]

OpenStudy (jfraser):

true. now just plug in everything you have

OpenStudy (anonymous):

ok, I got 50.31 as my answer

OpenStudy (jfraser):

that's what I get, GJ

OpenStudy (anonymous):

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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