# Consider the chemical reaction aC_2H_6 + bCO_2 + cH_2O--> dC_2H_5OH, where a,b,c and d are unknown positive integers. The reaction mush be balanced; that is, the number of atoms of each...

Consider the chemical reaction

where a,b,c and d are unknown positive integers. The reaction mush be balanced; that is, the number of atoms of each element must be the same before and after the reaction. While there are many possible choices for a,b,c and d that balance the reaction, it is customary to use the smallest possible integers. Balance this reaction.

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Hi, bleeeeeee,

Let's try this. First, for the C atoms, we would have:

**2a + b = 2d - equation #1**

Since they contribute 2 C atoms to the first compound, 1 to the CO2, and 2 to the last compound. Then, for the H atoms:

**6a +2c = 6d - equation #2**

Then, for the O atoms:

**2b + c = d - equation #3**

We can multiply equation #1by 3 on each side, getting:

**6a + 3b = 6d**

Subtracting equation #2 from this, we would get:

**3b - 2c = 0**

3b = 2c

**c = 3/2 b**

Given there are infinite values for the equation but the norm is to use the smallest integers we can, from here, we would see that if b = 2, then:

**c = 3/2 * 2 = 3**

So, we would have:

**b = 2, c = 3**

Plugging these back into equation #3, we would have:

2*2 + 3 = d

**d = 7**

Then, plugging into equation #1 (or #2):

2a + 2 = 2*7

2a + 2 = 14

2a = 12

**a = 6**

So:

**a = 6, b = 2, c = 3, d = 7**

Good luck, bleeeeeee. I hope this helps.

a(**C**_2**H**_6) + b(**CO**_2) + c(**H**_2**O**)--> d(**C**_2**H**_5**OH**)

First, **C** atoms, we have

2a + b = 2d (i)

For the **H** atoms, we have

6a +2c = 6d

3a+c=3d (ii)

For the **O** atoms, we have

2b + c = d (iii)

multiply equation (i) by 2 and subtract (iii) from it.

2(2a+b)-(2b+c)=2.2d-d

4a-c=3d (iv)

Now (ii)+(iv) ,we have

7a=6d

a=(6d)/7 (v)

substitute a from (v) ,in (i)

2(6d/7) + b = 2d

b=(2d)/7 (vi)

substitute b from (vi) in(iii)

2(2d/7) + c = d

c=(3d)/7 (vii)

i.e. a=(6d)/7 ,b=(2d)/7, c=(3d)/7

Since a,b,c are positive integers ,so if we choose d=7 then

a=6 ,b=2, c=3 .

Thus solution of the problem is

a=6,b=2,c=3 ,and d=7

**6a + 3b = 6d**

Subtracting equation #2 from this, we would get:

**3b - 2c = 0**

3b = 2c

**c = 3/2 b**