Linear substitution application

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Linear substitution application

Mathematics
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I need help balancing a chemical equation (now before you send me off to the chemistry section), the way to solve it involves math: I have to balance this equation: Cu + HNO3 = Cu(NO3)2 + NO + H2O I find that the answer is 3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O But I have no idea how to get there I found a wesite talking about using a linear substittuion, basically rewriting the equation like this a*Cu + b*HNO3 = c*Cu(NO3)2 + d*NO + e*H2O Since the same number of atoms exist on both sides, it becomes: Cu: a=c H: b=2e N: b=2c+d O: 3b=6c+d+e
@nincompoop @phi @TheSmartOne Any help please?
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Other answers:

are you short an equation? b/c there are 5 unknowns
How to solve for each unknown
@peachpi I was thinking the same thing.
Here is where I found the technique http://www.chemicalaid.com/articles.php/view/1/how-to-balance-chemical-equations
@TheĐ…martOne
need the matrix method to solve it???
I don't have 5 equations though
the example cleverly had enough eqautions to solve the problem.
I arrange it as \(x_1(Cu) +x_2(HNO_3)\rightarrow x_3(Cu(NO_3)_2)+x_4(NO)+x_5(H_2O)\) Left hand side Righ hand side Cu \(x_1\) = \(x_3\) Hydrogen \(x_2\) = \(2x_5\) Nitron \(2x_2\) = \(2x_3+x_4\) Oxygen \(3x_2\) = \(6x_3+x_4+x_5\)
Cu: a=c H: b=2e N: b=2c+d O: 3b=6c+d+e
a*Cu + b*HNO3 = c*Cu(NO3)2 + d*NO + e*H2O
Where do we go from there
It gives you: \(x_1-x_3=0\\x_2-2x_5=0\\2x_2-2x_3-x_4=0\\3x_2-6x_3-x_4-x_5=0\) hence the matrix is \[\left[\begin{matrix}1&0&-1&0&0&0\\0&1&0&0&-2&0\\0&2&-2&-1&0&0\\0&3&-6&-1&-1&0\end{matrix}\right]\]
When I do rref, I get -.25 -2 -.25 and -3.5
Take row reduce echelon form, you have \(x_1=(1/4)x_5\\x_2=2x_5\\x_3=(1/4)x_5\\x_4=(7/2)x_5\) hence if \(x_5= 4\) then \(x_1= 1\\x_2=8\\x_3=1\\s_4=14\)
Where did you ger that that x_1=1/4x_5
Now, replace all \(1(Cu) +8(HNO_3)\rightarrow 1(Cu(NO_3)_2)+14(NO)+4(H_2O)\)
That's not correct. It's 3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O
from the matrix, take row reduce echelon form, you have the first row 1 0 0 0 -1/4 0 that is x1 -1/4 x5 =0
oh ok
I may make mistake at somewhere, but the technique is that. Do it carefully and you can get the answer
Thank you very much
ok
It' in the matrix, you put for Nitrogen, 2x instead of just x
Yes!!!!!!!!!!! it is. My bad. I am sorry
then the rref is \(x_1=(3/4)x_5\\x_2= 2x_5\\x_3=(3/4)x_5\\x_4=(1/2)x_5\) if \(x_5=4\) then \(x_1=3\\x_2=8\\x_3=3\\x_4=2\) and the equation is the correct one as you had.
Thanks for the question!! I have an opportunity to recall my knowledge. :)

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