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balance `NaBH_4 + BF_3 -> NaBF_4 + B_2H_6`

balance `NaBH_4 + BF_3 -> NaBF_4 + B_2H_6`

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balance `NaBH_4 + BF_3 -> NaBF_4 + B_2H_6`

Postby Integr88 » Fri 01 April, 2011 23:01

How to balance `NaBH_4 + BF_3 -> NaBF_4 + B_2H_6`
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Re: balance `NaBH_4 + BF_3 -> NaBF_4 + B_2H_6`

Postby barnamah » Fri 01 April, 2011 23:10

to balance `NaBH_4 + BF_3 -> NaBF_4 + B_2H_6`
start with H, we have `H_6` on the right so we have to make H on the left 6 by multiplying by `3/2` because `3/2*4=6`

`3/2NaBH_4 + BF_3 -> NaBF_4 + B_2H_6`

Now H is balanced by we have changed Na so on the right we add `3/2` for the Na and we have:
`3/2NaBH_4 + BF_3 -> 3/2NaBF_4 + B_2H_6` Now Na is balanced

Now on the right B is changed and we have `2+3/2` B on the right so we have add a 2 one left for B and we have:
`3/2NaBH_4 + 2BF_3 -> 3/2NaBF_4 + B_2H_6`

Now if we look at our equation, it is balanced but we have fraction. To make it whole number and because we have 2 at the denominator, we can multiply the whole equation by 2 and we have:

`2*3/2NaBH_4 + 2*2BF_3 ->2* 3/2NaBF_4 +2* B_2H_6`
simplify and we have:

`3NaBH_4 + 4BF_3 ->3NaBF_4 +2B_2H_6`
Explore and know. That is asked.
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