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
@Jamierox4ev3r
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 ?
Not quite, unfortunately
hmmmm....
This is also another double replacement reaction. So: \(Li + H_2O~ -> LIOH + H_2\)
LIOH+H2?
yup!
So, based on how I helped you on the previous question, can you attempt to balance the equation for me? I can guide you
li= 1 li = 1 h= 2 h= 3 0= 1 0=1
Good! So we can see discrepancies in the amounts of hydrogen, mainly.
So where would you add subscripts to fix this?
Li0+1h2
Huh?
oh sorry, I meant coefficients.
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
what happens if we have 2 LiOH? @rebeccaxhawaii
li= 3 0 = 3 H= 3
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.
This is really important to understand if you want to balance equations on your own
oh okay
\(Li + H_2O~ -> \color{red}{2}LIOH + H_2\) *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!
Can you figure out where the other coefficients need to go then?
2Li+2H2O
bingo!
2Li+2H2O => 2LIOH+H2
\(\color{red}{2}Li + \color{red}{2}H_2O~ -> \color{red}{2}LiOH + H_2\) You are correct
alright awesome! THANK YOU SO MUCHHHHHHHHHH
of course, anytime :)
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