Ag 2HNO3 = AgNO3 NO2 H2O

(Unless there are side reactions I'm missing)

I've got 24.4g of silver to dissolve, which means I need 28.5g of HNO3.

My nitric is ~30%, so I need 95g of my acid to get that much HNO3 (and 66.5g of H2O). I'll probably use up to 100g, adding extra as needed with an addition funnel, just in case it's not as strong as my rough density calculation predicted.

The reaction will produce ~38.43g of silver nitrate and ~10g of NO2 that I'll scrub through a sodium hydroxide solution.

66.5g (plus about 5g from the reaction) is plenty of water to dissolve the produced silver nitrate, so I shouldn't have to worry about passivation, as long as I did my math properly. I'll also have it on reasonably strong stirring to keep the solution moving and to maybe recycle some of the NO2 back into acid.