Question

Cellular respiration transforms glucose and oxygen into carbon dioxide, water, and energy. C6H12O6 + 6O2 → 6CO2 + ?H2O + Energy Based on the law of conservation of matter, what is the missing coefficient for water? (1 point) Responses 4 4 2 2 6 6 8

Answer 1

To determine the missing coefficient for water in the cellular respiration equation \(C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + ?H_2O + \text{Energy}\), we can use the law of conservation of matter, which states that the number of atoms for each element must be the same on both sides of the equation.

Here's the breakdown of the existing elements:

1. **Glucose**: \(C_6H_{12}O_6\)
- Carbon (C): 6
- Hydrogen (H): 12
- Oxygen (O): 6

2. **Oxygen (input)**: \(6O_2\)
- O: 12 (since \(6 \times 2 = 12\))

On the right side of the equation:
1. **Carbon dioxide**: \(6CO_2\)
- C: 6
- O: 12 (since \(6 \times 2 = 12\))

Let \(?H_2O\) be represented as \(xH_2O\):
- In \(xH_2O\):
- O: \(x\)
- H: \(2x\)

Now, we can set up the equation based on the conservation of matter.

**Balance for Carbon:**
- Left: 6 C from \(C_6H_{12}O_6\)
- Right: 6 C from \(6CO_2\)

**Balance for Oxygen:**
- Left: \(6\) O from \(C_6H_{12}O_6\) + \(12\) O from \(6O_2\) = \(18\) O total
- Right: \(12\) O from \(6CO_2\) + \(x\) O from \(xH_2O\) = \(12 + x\)

Setting the equations equal for oxygen:
\[
18 = 12 + x
\]
Solving for \(x\):
\[
x = 18 - 12 = 6
\]

So the equation is \(C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}\).

Thus, the missing coefficient for water is **6**.