See if you can Balance This Chemical Reaction!?!?

Question

Cu(2)S + HNO(3) ---> Cu(NO(3))2 + CuSO(4) + NO(2) + H(2)0

and name it if you can in their real chemical names. I have been trying forever, and I cant figure it out! PLZ HELP! The numbers in the perenthesis are subscripts. So please try and balance it!...Thx

Answer 1

I tried assigning variable multipliers to each of the chemicals and writing the associated equations. The variables are "a" through "f" in order in which they are writen in your equation.

I get the following set of linear equations:

(Cu) 2a = c + d
(S) a = d
(H) b = 2f
(N) b = 2c + e
(O) 3b = 6c + 4d + 2e + f

Since there are 6 unknowns and only 5 equations start by arbitrarily setting a = 1
then d = 1 also.
And c = 1

So we now have
b = 2f
b = 2 + e
3b = 10 + 2e + f

substituting for e and f in the 3rd equation we get
3b = 10 + 2(b-2) + (b/2)
3b -2b - b/2 = 10 -4
b/2 = 6
b = 12
then f = 6
and e = 10

So your balanced equation should be
Cu(2)S + 12HNO(3) --> Cu(NO(3))2 + CuSO(4) + 10NO(2) + 6H(2)O

Answer 2

nicely for first one, you will need at the start left area first we could take CaCl2 this implies , you will desire to make certain one million Ca , and a pair of Cl on desirable area. Ca is then returned dont replace the multiplying area of any factor that has Ca in it. you will desire to placed 2 * ( NaCl) on desirable area that makes Cl numbers even. nicely once you end with Cl, you will desire to learn Na for the reason which you alter its multiplying element on the left area , you have one Na and on the main appropriate area of the equation you have 2 Na so placed 2 till now NaHCo3. then ultimately verify the HCO3 numbers on the two aspects it's going to be like this CaCl2 + 2 * NaHCO3 -----> Ca ( HCO3)2 + 2 * NaCl for 2nd equation, there is not any replace for the reason that is in balanced variety.