Write the reaction as a CO2 + b H2O → c C6H12O6 + d O2 where a, b, c, and d are meant to be integers adjusted so that the number of each kind of atoms balance on each side. For this reaction, we have three kinds of atoms that must balance: carbon (C), hydrogen (H), and oxygen (O). Write three equations for a, b, c, and d that, if satisfied, will guarantee that the number of atoms of each kind are the same on each side of the reaction.

a = 6, b = 6, c = 1 and d = 6

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Explanation:

a CO₂ + b H₂O → c C₆H₁₂O₆ + d O₂

In the balance for Carbon (C)

a = 6c

In the balance for Oxygen (O)

2a + b = 6c + 2d

In the balance for Hydrogen (H)

2b = 12 c

b = 6c

To solve, we bring the equations together

a = 6c (eqn 1)

2a + b = 6c + 2d (eqn 2)

b = 6c (eqn 3)

Substituting for b and a in (eqn 2)

2a + b = 6c + 2d

2(6c) + 6c = 6c + 2d

12c + 6c = 6c + 2d

2d = 12c

d = 6c

Let c = 1

a = 6c = 6

b = 6c = 6

d = 6c = 6

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

trigonal pyramidal

tetrahedral

trigonal planar

bent