For the reaction 2H2O2 -> 2H2O + O2, what is the ratio of the rate of disappearance of H2O2 to the rate of formation of O2?

The given reaction is:

$$2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$$

For every 2 moles of $\text{H}_2\text{O}_2$ that disappear, 1 mole of $\text{O}_2$ is formed.
This means that the rate of disappearance of $\text{H}_2\text{O}_2$ is twice the rate of formation of $\text{O}_2$.

Mathematically, the relationship between the rates is:

$$-\frac{1}{2}\frac{d[\text{H}_2\text{O}_2]}{dt} = \frac{d[\text{O}_2]}{dt}$$

Where $-\frac{d[\text{H}_2\text{O}_2]}{dt}$ is the rate of disappearance of $\text{H}_2\text{O}_2$, and $\frac{d[\text{O}_2]}{dt}$ is the rate of formation of $\text{O}_2$.

We need to find the rate of disappearance of $\text{H}_2\text{O}_2$ with respect to the rate of formation of $\text{O}_2$. This is the ratio:

$$\frac{-\frac{d[\text{H}_2\text{O}_2]}{dt}}{\frac{d[\text{O}_2]}{dt}}$$

From the relationship in Step 2, multiply both sides by 2:

$$-\frac{d[\text{H}_2\text{O}_2]}{dt} = 2 \times \frac{d[\text{O}_2]}{dt}$$

Now, divide both sides by $\frac{d[\text{O}_2]}{dt}$:

$$\frac{-\frac{d[\text{H}_2\text{O}_2]}{dt}}{\frac{d[\text{O}_2]}{dt}} = 2$$

The ratio is 2.

The correct option is A.
The rate of disappearance of $\text{H}_2\text{O}_2$ with respect to the rate of formation of $\text{O}_2$ gas is $\boxed{\text{A. 2}}$.