The Timpson Lyceum

1. (2 pts) Balance the chemical equation by finding the smallest positive integer coefficients:

$x_1\,\mathrm{Fe} + x_2\,\mathrm{O_2} \longrightarrow x_3\,\mathrm{Fe_2O_3}.$

Solution

1. Match atoms on each side. Iron: $x_1 = 2x_3$ . Oxygen: $2x_2 = 3x_3$ . Take $x_3 = 2$ : then $x_1 = 4$ and $x_2 = 3$ . The balanced equation is

$4\,\mathrm{Fe} + 3\,\mathrm{O_2} \longrightarrow 2\,\mathrm{Fe_2O_3}.$

2. (3 pts) Traffic (cars per hour) flows through four one-way intersections $A, B, C, D$ . Conservation of flow (in $=$ out) at each intersection gives

$\begin{aligned} \text{$A$:}\quad x_1 + 40 &= x_2 + 30 \\ \text{$B$:}\quad x_2 + 50 &= x_3 + 60 \\ \text{$C$:}\quad x_3 + 20 &= x_4 + 40 \\ \text{$D$:}\quad x_4 + 30 &= x_1 + 10. \end{aligned}$

Solve the system, describe the solution set, and determine the minimum possible value of $x_1$ given that all flows are nonnegative.

Solution

2. Rewrite each equation as $\text{(net)}$ :

$\begin{aligned} x_1 - x_2 &= -10 \\ x_2 - x_3 &= 10 \\ x_3 - x_4 &= 20 \\ x_4 - x_1 &= -20. \end{aligned}$

The four equations sum to $0 = 0$ , so one is redundant. Let $x_4 = t$ be free. Then $x_3 = 20 + t$ , $x_2 = 10 + x_3 = 30 + t$ , and $x_1 = -10 + x_2 = 20 + t$ (this also satisfies the fourth equation: $t - (20+t) = -20$ ). So

$\mathbf{x} = \begin{bmatrix} 20 \\ 30 \\ 20 \\ 0 \end{bmatrix} + t\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \qquad t \ge 0.$

All flows are nonnegative iff $t \ge 0$ (the binding constraint is $x_4 = t \ge 0$ ). The minimum value of $x_1 = 20 + t$ is therefore $\boxed{20}$ , achieved at $t = 0$ .

3. (3 pts) Balance the combustion of propane:

$x_1\,\mathrm{C_3H_8} + x_2\,\mathrm{O_2} \longrightarrow x_3\,\mathrm{CO_2} + x_4\,\mathrm{H_2O}.$

Solution

3. Carbon: $3x_1 = x_3$ . Hydrogen: $8x_1 = 2x_4$ , so $x_4 = 4x_1$ . Oxygen: $2x_2 = 2x_3 + x_4 = 6x_1 + 4x_1 = 10x_1$ , so $x_2 = 5x_1$ . Take $x_1 = 1$ : $x_2 = 5$ , $x_3 = 3$ , $x_4 = 4$ :

$\mathrm{C_3H_8} + 5\,\mathrm{O_2} \longrightarrow 3\,\mathrm{CO_2} + 4\,\mathrm{H_2O}.$

4. (3 pts) A nutritionist blends three foods. Per unit serving, the grams of protein, carbohydrate, and fat are:

$\text{Food A}: (2, 1, 0), \quad \text{Food B}: (1, 3, 1), \quad \text{Food C}: (0, 1, 2),$

where each triple is (protein, carbohydrate, fat). Find the numbers of servings $a, b, c$ that provide exactly $7$ g protein, $12$ g carbohydrate, and $5$ g fat.

Solution

4. The constraints give the system

$\begin{aligned} 2a + b &= 7 \\ a + 3b + c &= 12 \\ b + 2c &= 5. \end{aligned}$

From the first, $b = 7 - 2a$ ; from the third, $c = \tfrac{5 - b}{2} = \tfrac{5 - (7 - 2a)}{2} = a - 1$ . Substitute into the second: $a + 3(7 - 2a) + (a - 1) = 12 \Rightarrow a + 21 - 6a + a - 1 = 12 \Rightarrow -4a + 20 = 12 \Rightarrow a = 2$ . Then $b = 3$ and $c = 1$ . Answer: $2$ servings of A, $3$ of B, $1$ of C. (Check: protein $4+3 = 7$ , carbohydrate $2+9+1 = 12$ , fat $3+2 = 5$ .)

5. (3 pts) Balance the photosynthesis reaction:

$x_1\,\mathrm{CO_2} + x_2\,\mathrm{H_2O} \longrightarrow x_3\,\mathrm{C_6H_{12}O_6} + x_4\,\mathrm{O_2}.$

Solution

5. Carbon: $x_1 = 6x_3$ . Hydrogen: $2x_2 = 12x_3$ , so $x_2 = 6x_3$ . Oxygen: $2x_1 + x_2 = 6x_3 + 2x_4$ ; substituting, $12x_3 + 6x_3 = 6x_3 + 2x_4 \Rightarrow 2x_4 = 12x_3 \Rightarrow x_4 = 6x_3$ . Take $x_3 = 1$ :

$6\,\mathrm{CO_2} + 6\,\mathrm{H_2O} \longrightarrow \mathrm{C_6H_{12}O_6} + 6\,\mathrm{O_2}.$

6. (3 pts) State whether each of the following is true or false, and justify each answer.

(a) The coefficients that balance a chemical equation are found by solving a homogeneous linear system.

(b) A network-flow system always has a unique solution.

(c) In a flow network, the equation at each node states that total flow in equals total flow out.

Solution

6.

(a) True. Conservation of each element gives one equation with zero on the right (atoms in $=$ atoms out), so the coefficients solve a homogeneous system; the balanced equation comes from a nontrivial solution scaled to smallest integers.

(b) False. A flow network typically yields a dependent system with a free variable, so the solution is a one- (or more-) parameter family, not unique, unless enough branch flows are measured.

(c) True. Conservation of flow at a node is exactly the statement that total inflow equals total outflow.

7. (3 pts) Explain why the node equations of a closed flow network (like Problem 2) are always dependent — their sum is $0 = 0$ — so the system has at least one free variable. Interpret what happens to the solution when one branch flow is measured directly.

Solution

7. Every branch of a closed network leaves one node and enters another, so its flow appears once with a $+$ sign (as an inflow) and once with a $-$ sign (as an outflow) across the node equations. Adding all node equations, each branch variable cancels and every measured inflow cancels against its outflow, leaving $0 = 0$ . Thus the equations are linearly dependent: one is redundant, so with $n$ unknown branch flows there are at most $n-1$ independent equations, forcing at least one free variable. Measuring one branch flow directly supplies the missing equation, pinning down the free parameter and (typically) determining all remaining flows uniquely.

8. (4 pts ••) Balance the reaction

$x_1\,\mathrm{KMnO_4} + x_2\,\mathrm{HCl} \longrightarrow x_3\,\mathrm{KCl} + x_4\,\mathrm{MnCl_2} + x_5\,\mathrm{H_2O} + x_6\,\mathrm{Cl_2},$

finding the smallest positive integer coefficients.

Solution

8. Balance each element:

$\begin{aligned} \text{K:}\ & x_1 = x_3 & \text{Mn:}\ & x_1 = x_4 \\ \text{O:}\ & 4x_1 = x_5 & \text{H:}\ & x_2 = 2x_5 \\ \text{Cl:}\ & x_2 = x_3 + 2x_4 + 2x_6. & & \end{aligned}$

From K and Mn, $x_3 = x_4 = x_1$ . From O, $x_5 = 4x_1$ ; from H, $x_2 = 2x_5 = 8x_1$ . Substitute into the Cl balance: $8x_1 = x_1 + 2x_1 + 2x_6 \Rightarrow 2x_6 = 5x_1 \Rightarrow x_6 = \tfrac{5}{2}x_1$ . Take $x_1 = 2$ to clear the fraction: $x_2 = 16$ , $x_3 = 2$ , $x_4 = 2$ , $x_5 = 8$ , $x_6 = 5$ . The balanced equation is

$2\,\mathrm{KMnO_4} + 16\,\mathrm{HCl} \longrightarrow 2\,\mathrm{KCl} + 2\,\mathrm{MnCl_2} + 8\,\mathrm{H_2O} + 5\,\mathrm{Cl_2}.$

(Chlorine check: right side $2 + 4 + 10 = 16$ = left side.) $\blacksquare$