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Chemistry 8 Online
rebeccaxhawaii (rebeccaxhawaii):

Write out the balanced equation for the reaction that occurs when Li and H2O react together. You do not need to make your subscripts smaller; just write them out as regular numbers. For example: H2O

rebeccaxhawaii (rebeccaxhawaii):

@Jamierox4ev3r

OpenStudy (zyi6):

Well, Lithium has a +1 charge on it at all times, so we can write the reaction product as this: \[Li _{2}HO\] I think this is the correct way to write it. Am I wrong @Jamierox4ev3r ?

HanAkoSolo (jamierox4ev3r):

Not quite, unfortunately

OpenStudy (zyi6):

hmmmm....

HanAkoSolo (jamierox4ev3r):

This is also another double replacement reaction. So: \(Li + H_2O~ -> LIOH + H_2\)

rebeccaxhawaii (rebeccaxhawaii):

LIOH+H2?

HanAkoSolo (jamierox4ev3r):

yup!

HanAkoSolo (jamierox4ev3r):

So, based on how I helped you on the previous question, can you attempt to balance the equation for me? I can guide you

rebeccaxhawaii (rebeccaxhawaii):

li= 1 li = 1 h= 2 h= 3 0= 1 0=1

HanAkoSolo (jamierox4ev3r):

Good! So we can see discrepancies in the amounts of hydrogen, mainly.

HanAkoSolo (jamierox4ev3r):

So where would you add subscripts to fix this?

rebeccaxhawaii (rebeccaxhawaii):

Li0+1h2

HanAkoSolo (jamierox4ev3r):

Huh?

HanAkoSolo (jamierox4ev3r):

oh sorry, I meant coefficients.

HanAkoSolo (jamierox4ev3r):

But in any case: \(Li + H_2O~ -> LIOH + H_2\) *1 Li *1 Li *2 H \(\color{red}{ *3 H}\) *1 O *1 O We have the right amount of O and Li already However, since H is off, we can add a coefficient of 2 in front of LiOH in order to start fixing this

HanAkoSolo (jamierox4ev3r):

what happens if we have 2 LiOH? @rebeccaxhawaii

rebeccaxhawaii (rebeccaxhawaii):

li= 3 0 = 3 H= 3

HanAkoSolo (jamierox4ev3r):

not quite. 2 LiOH just gives us the following: *2 Li *2 O *2 H This is all just multiplication, so it's the current amount x 2.

HanAkoSolo (jamierox4ev3r):

This is really important to understand if you want to balance equations on your own

rebeccaxhawaii (rebeccaxhawaii):

oh okay

HanAkoSolo (jamierox4ev3r):

\(Li + H_2O~ -> \color{red}{2}LIOH + H_2\) *1 Li *2 Li *2 H *4 H *1 O *2 O So this is what we have so far. But noticed how everything in the reactants just needs to be doubled!

HanAkoSolo (jamierox4ev3r):

Can you figure out where the other coefficients need to go then?

rebeccaxhawaii (rebeccaxhawaii):

2Li+2H2O

HanAkoSolo (jamierox4ev3r):

bingo!

rebeccaxhawaii (rebeccaxhawaii):

2Li+2H2O => 2LIOH+H2

HanAkoSolo (jamierox4ev3r):

\(\color{red}{2}Li + \color{red}{2}H_2O~ -> \color{red}{2}LiOH + H_2\) You are correct

rebeccaxhawaii (rebeccaxhawaii):

alright awesome! THANK YOU SO MUCHHHHHHHHHH

HanAkoSolo (jamierox4ev3r):

of course, anytime :)

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