1. We analyze the equilibrium:

$2\,\mathrm{SO_2}(g)+\mathrm{O_2}(g) \rightleftharpoons 2\,\mathrm{SO_3}(g), \qquad \Delta H=-190\,\mathrm{kJ}$

Since $\Delta H<0$, the forward reaction is exothermic.

2. Check each factor using Le Chatelier’s principle.

A. Increasing temperature

For an exothermic reaction, heat behaves like a product. Increasing temperature shifts equilibrium to the left, decreasing $\mathrm{SO_3}$ yield.

• Does not increase yield

B. Increasing pressure

Count gaseous moles:

• Left side: $2+1=3$ moles
• Right side: $2$ moles

Increasing pressure favors the side with fewer moles of gas, i.e. the right side.

• Increases yield of $\mathrm{SO_3}$

C. Adding more $\mathrm{SO_2}$

Increasing concentration of a reactant shifts equilibrium to the right to consume added reactant.

• Increases yield of $\mathrm{SO_3}$

D. Adding more $\mathrm{O_2}$

Similarly, adding another reactant shifts equilibrium to the right.

• Increases yield of $\mathrm{SO_3}$

E. Addition of catalyst

A catalyst speeds up both forward and backward reactions equally. It helps reach equilibrium faster but does not change equilibrium composition.

• Does not increase yield

3. Therefore, the factors that increase the equilibrium yield of $\mathrm{SO_3}$ are:

$\boxed{B,\ C,\ D}$

So, the number of such factors is

$\boxed{3}$