3. Chemistry In chemistry we can use a system of linear equations to determine t

Question

3. Chemistry In chemistry we can use a system of linear equations to determine the coefficients neces- sary to balance a chemical equation. For example, the complete combustion of propene (C3Hs) with oxygen (O2) results in carbon dioxide (CO2) and water Because the quantity of each element is conserved, we must have the same quantities of carbon (C), oxygen (o) and hydrogen (H) on each side of the following chemical where the coefficients a, b,c, d are how many of each molecule is present. (a) the equation describing the conserved quantity of each type of chemical. You will have 3 equations unknowns. The equation for conserved oxygen atoms is given as an example: (b) Write this system in matrix form

Explanation / Answer

a(C3H8) + bO2 = c(CO2) + d(H2O)

Part (a)

Equation for carbon C:

3a = c

3a - c = 0 ...... (eqn 1)

Equation for H:

8a = 2d

8a - 2d = 0 .... (eqn 2)

Equation for O:

2b = 2c + d

2b - 2c - d = 0 .... (eqn 3)

Number of equations = 3

Number of variables = 4 (a, b, c, d)

Part (b)

equation in matrix form will be (see the image attached):

part(c)

keep c as the variable = k

from eqn (1)

3a = c

3a = k

a = k/3

from eqn (2)

8a = 2d

d = 4a

d = 4*(k/3) = 4k/3

from equation (3)

b = (2c+d)/2

b = ((2*k)+ (4k/3))/2

b = 5k/3

part (d)

2(C3H8) + bO2 = c(CO2) + d(H2O) + e(CO)

solving for parameter b, we get

Equation for C:

6 = c + e...... (eqn 1 )

Equation for O:

2b = 2c + d + e ..... (eqn 2)

Equation for H:

16 = 2d

d = 8 ... eqn(3)

now,

eqn (2) becomes

2b = 2c + 8 + e

from eqn(1)

c = 6 - e

put value of C in eqn (2)

2b = 2(6 - e) + 8 + e

hence, e = 20 - 2b

hence, c = 2b - 14

the ranges for parameters will be

for e > 0

20 - 2b > 0

b <10

ranges for c

c > 0

2b - 14 > 0

b > 7

Hence, 7<b<10 is the range

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