- #1
v3ra
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Homework Statement
(a)H2SO4 + (b)NaHCO3 -> (x)Na2SO4 + (y)CO2 + (z)H2O
I gave each element it`s own equation:
(1) For hydrogen – 2a + b = 2z
(2) For sulphur – a = x
(3) For oxygen – 4a + 3b = 4x + 2y + z
(4) For sodium – b = 2x
I then assigned a number to letter 'a'. My book says to begin with the number 12 because it is the most conventionally used.
I began with equation (2) since it had the least amount of variables and proceeded in this fashion.
a = 12
(2)
a = x
x = 12
(4)
b = 2x
b = 2(12)
b = 24
(1)
2a + 2b = 2z
2(12) + 2(24) = 2z
24 + 48 = 2z
72 = 2z
z = 36
(3)
4a + 3b = 4x + 2y + z
4(12) + 3(24) = 4(12) + 2y + 36
48 + 72 = 48 + 2y + 36
120 = 84 + 2y
36 = 2y
y = 18
I then divided each of the values by a common divider, in this case '6' in order to get the lowest possible coefficient. I ended up with:
a = 2
b = 4
x = 2
y = 3
z = 6
When I add these to my original formula, it is not balanced:
(2)H2SO4 + (4)NaHCO3 -> (2)Na2SO4 + (3)CO2 + (6)H2O
I also tried dividing by 12, 4 and 2 and I am still left with an unbalanced formula.
The book says that when this happens, I must instead assign the value of 1 to 'a', I did this and got the following results:
a = 1
(2)
a = x
x = 1
(4)
b = 2x
b = 2(1)
b = 2
(1)
2a + 2b = 2z
2(1) + 2(2) = 2z
2 + 4 = 2z
6 = 2z
z = 3
(3)
4a + 3b = 4x + 2y + z
4(1) + 3(2) = 4(1) + 2y + 3
4 + 6 = 4 + 2y + 3
10 = 7 + 2y
3 = 2y
y = 1.5
As you can see, 'y' is left at 1.5 which is not a whole number and therefore unacceptable. Can someone please tell me what I am missing?