Battery power required for a 2HP Motor AC at 1800 rpm, 56Hz?
The maths


  2.3 kW / .85 power factor = 2.7 kW / 220 V = 12.27 A.


Yes, at 220 V.



At 12 V ...

2.3 kW at 12 V: \$ I = \frac {P}{V} = \frac {2300}{12} \approx 200~A \$.

For 5 hour run time you would need 1000 Ah capacity with a 100% efficient inverter. Assuming you could find an 80% efficient inverter and you decide to only discharge the batteries by 75% to prolong their life then required battery capacity:

$$ Ah = \frac {1000}{80\% \times 75\%} = \frac {1000}{0.8 \times 0.75} = 1666~Ah $$

As you can see, your 50 Ah is not going to work.



Selecting battery voltage

When designing high-powered inverters and UPS the standard technique is to use a high voltage battery bank rather than a 12 V bank. This has several advantages:


The current is lower.
Conductor sizes can be reduced.
Power loss through the switching devices (e.g., transistors) is less because of the reduced current. This reduces the heatsinking requirements.




Basic inverter calculations:

$$ P_{OUT} = \eta P_{IN} $$

where \$ \eta \$ is the inverter efficiency.