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Consider the chemical reaction aC_2H_6 + bCO_2 + cH_2O--> dC_2H_5OH,   where a,b,c...

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bleeeeee | Student | Honors

Posted March 13, 2013 at 11:11 PM via web

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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 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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steveschoen | College Teacher | (Level 3) Assistant Educator

Posted March 14, 2013 at 11:11 PM (Answer #1)

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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.

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pramodpandey | College Teacher | Valedictorian

Posted March 17, 2013 at 8:10 AM (Answer #2)

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a(C_2H_6) + b(CO_2) + c(H_2O)--> d(C_2H_5OH)

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

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