Fe(OH)3(S) +3HNO3(aq)----->Fe(NO3)3(aq) + 3H20(aq)
M(Fe(OH)3)=56+48+3=107; M(HNO3)= 48+14+1=63
n(Fe(OH)3)=5.4/107=0.05; n(HNO3)=2.6/63=0.04
n(Fe(OH)3):n(HNO3)=1:3, which means that the HNO3 should be three times (molar) than the Fe(OH)3, but you can see that it is, actually, even less than the Fe(OH)3, meaning that HNO3 is the limiting reagent and the amount of Fe(OH)3 which is going to react with HNO3 is 0.04/3=0.013 i.e. 0.05-0.013=0.037 mol Fe(OH)3 is left after the completion.
Just in case you can convert it into mass, but I suppose this is enough.