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Chemistry 8 Online
OpenStudy (anonymous):

can someone explain this to me ?

OpenStudy (anonymous):

OpenStudy (anonymous):

where does the 1mol Li come from in 1mol Li/6.94 g Li?

OpenStudy (anonymous):

im not trying to cheat I'm trying to understand everything, I'm just a little lost

OpenStudy (abb0t):

Molar mass is grams per 1 unit mole. That is the approximate mass of 1 atom of Lithium.

OpenStudy (anonymous):

alright thanks, that makes sense. what about the equation 1mol H2/2mol Li ?

OpenStudy (anonymous):

@abb0t one other question?

OpenStudy (abb0t):

You refer to the balanced chemical equation. For example, you have 2 Li + H\(_2\)\(\rightarrow\) 2LiH\(_2\) It's a 2:1 ratio, that means you have TWO lithium ions, or 2 moles, and 1 hydrogen. So that is why you use 2 over 1. The lithium goes on the bottom because it cancels units nicely: \(\sf \color{red}{1~\cancel{mole~Li}\times \frac{1~mol~H_2}{2~\cancel{mol~Li}}}\) = mole of H\(_2\). It is the same as fractions from algebra.

OpenStudy (anonymous):

alright thank you can you tell me if I'm doing this equation right @abb0t 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. 2 Li (s) + 2 H2O (l) yields 2 LiOH (aq) + H2 (g)

OpenStudy (anonymous):

34.2 g L x 1mol Li/ 2.94 g Li x 1mole x 1mole H2/ 2mol Li = 5.8

OpenStudy (abb0t):

Well, first, you need to write down the complete balanced chemical equation! That is always the first step when doing these types of problems. Can you do that?

OpenStudy (anonymous):

and this would be a 2: 1 ratio as well from what you just told me

OpenStudy (anonymous):

um I think lol sorry i haven't done this in a year, isn't 2 Li (s) + 2 H2O (l) yields 2 LiOH (aq) + H2 (g) the complete equation ?

OpenStudy (anonymous):

@abb0t

OpenStudy (anonymous):

n(H2)=n(Li)2 n(Li)=m(Li)M(Li)=34.2g6.94gmol=4.927mol

OpenStudy (anonymous):

isn't balancing it = 34.2 g Li x 1mol Li/ 2.94 g Li x 1mole x 1mole H2/ 2mol Li

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