## anonymous one year ago Linear substitution application

1. anonymous

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

2. anonymous

@nincompoop @phi @TheSmartOne Any help please?

3. sparrow2

what is the question?

4. anonymous

are you short an equation? b/c there are 5 unknowns

5. anonymous

How to solve for each unknown

6. anonymous

@peachpi I was thinking the same thing.

7. anonymous

Here is where I found the technique http://www.chemicalaid.com/articles.php/view/1/how-to-balance-chemical-equations

8. Elsa213

@TheЅmartOne

9. Loser66

need the matrix method to solve it???

10. anonymous

I don't have 5 equations though

11. anonymous

the example cleverly had enough eqautions to solve the problem.

12. Loser66

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 $$3x_2$$ = $$6x_3+x_4+x_5$$

13. anonymous

Cu: a=c H: b=2e N: b=2c+d O: 3b=6c+d+e

14. anonymous

a*Cu + b*HNO3 = c*Cu(NO3)2 + d*NO + e*H2O

15. anonymous

Where do we go from there

16. Loser66

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 $\left[\begin{matrix}1&0&-1&0&0&0\\0&1&0&0&-2&0\\0&2&-2&-1&0&0\\0&3&-6&-1&-1&0\end{matrix}\right]$

17. anonymous

When I do rref, I get -.25 -2 -.25 and -3.5

18. Loser66

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 $$x_1= 1\\x_2=8\\x_3=1\\s_4=14$$

19. anonymous

Where did you ger that that x_1=1/4x_5

20. Loser66

Now, replace all $$1(Cu) +8(HNO_3)\rightarrow 1(Cu(NO_3)_2)+14(NO)+4(H_2O)$$

21. anonymous

That's not correct. It's 3Cu + 8HNO3 = 3Cu(NO3)2 + 2NO + 4H2O

22. Loser66

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

23. anonymous

oh ok

24. Loser66

I may make mistake at somewhere, but the technique is that. Do it carefully and you can get the answer

25. anonymous

Thank you very much

26. Loser66

ok

27. anonymous

It' in the matrix, you put for Nitrogen, 2x instead of just x

28. Loser66

Yes!!!!!!!!!!! it is. My bad. I am sorry

29. Loser66

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.

30. Loser66

Thanks for the question!! I have an opportunity to recall my knowledge. :)