To determine the moles of cations in 1.30 mol of K2SO4, we need to consider the stoichiometry of the compound. K2SO4 consists of two K+ ions (cations) and one SO42- ion (anion).

In 1 mole of K2SO4, there are 2 moles of K+ ions. Therefore, in 1.30 mol of K2SO4, there will be:

$$2 \ mol \ K^+ \ ions \ / 1 \ mol \ K_2SO_4 \times 1.30 \ mol \ K_2SO_4 = 2.60 \ mol \ K^+$$

Additionally, there are 1 mole of SO42- ions in 1 mole of K2SO4. However, since we are interested in the total moles of cations, we will not count the moles of anions.

Therefore, in 1.30 mol of K2SO4, there are 2.60 mol of cations.