I want to use pcb relay to control(on/off) the submersible motor-pump of 5 hp with MCU control.

for that, efficiency = 85%, pf=80%, at 230v, the Continuous current(I) will be,

Input=o/p / efficieny = 5/0.85 = 5.8 ==6 hp x 746 =4.476 kw.

I= 4.476/(230x0.8) = 25A running current.

--As the starting Amp of an Induction motor are 3 to 5 times the running current/Full load current FLA.

--So need Power relay which will withstand for such starting Amp (ie 25A x 5 = 125 Amp).

IS the relay model "JQX-15F (T90)" is sufficient to withstand and longer life. The datasheet is also attached here.
From the data sheet,
I=30A, Contact materials is AgCdo used. But the withstand current not found here in the datasheet.

Please suggest ,
How do i calculate/find the withstand current for that amount of starting current from this datashhet??
Is my all above calculation correct ?