3) q = 3, r = 12, s = 3, t = 6

There are a total of 12 Hydrogen atoms on the RHS and that is unchangeable.

Hydrogen is only present in HNO₃ on the LHS.

Therefore, to balance the Hydrogen atoms on both sides, ** r = 12**.

Copper is present on its own on LHS and in Cu(NO₃)₂ on RHS.

So __q = s__

Let us form algebraic equations.

**For Copper,**

__q = s__

**For Oxygen,**

3r = 6s + 6 + 2t

36 = 6s + 6 + 2t

__30 = 6s + 2t__ —— (1)

**For Hydrogen,**

__r = 12
__

**For Nitrogen,**

r = 2s + t

Since r = 12

__12 = 2s + t__ —— (2)

Using simultaneous equations,

6s + 2t = 30

(2s + t = 12) x 2

6s + 2t = 30

4s + 2t = 24

2s = 6

__s = 3__

Since (q = s) and (s = 3),

__q = 3__

Using equation (2),

12 = 2s + t

t = 12 – 2s

t = 12 – 2(3)

t = 12 – 6

__t = 6__q = 3, r = 12, s = 3, t = 6