(I got 13.25 grams of Na2CO3 would be needed to react with 0.250 moles HCl by multiplying 0.25 and 105.99g (molar mass of Na2CO3) - not sure if this is correct

Here, and in other problems, you are just juggling numbers and you don't even pay attention to what you are doing. In most cases you start right, but you ignore the stoichiometric coefficient in the final step.

0.25 times 105.99 is not 13.25

Accidentally, the answer you gave (13.25 g of Na

_{2}CO

_{3}) is correct, even if it doesn't follow your calculations.