Converting fuel level sender reading
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. A constant current source of 166 mA will give an output of 0.33 V at 2 &Omega; and 5 V at 30 &Omega;.

A constant current source prevents variations in fuel gauge readings at varying engine revolutions. A typical constant current source is shown in Figure 1. How it works:


R3 draws base current from Q2 turning it on.
As the current through Q2 rises the voltage drop across R2 increases and turn on Q1. This happens at about 0.75 V.
Q1 then turns on pulling the base of Q2 towards +12 V and preventing Q2 turning on any further. The circuit settles down at \$ I = \frac {0.7} {R_2} = \frac {0.75} {4.5} = 166 \ \text {mA} \$.


There are a few things to watch out for:


Q2 will have up to 13 V across it when the engine is revving high. \$ P_{Q2} = V_{Q2}I_{Q2} = 13 \times 0.166 = 2.1 \ \text W \$ so it will require heatsinking.
Beware that as drawn the voltage to your 0 - 5 V ADC will get 14 V if the sensor is ever disconnected. You would want to add some protection there to prevent frying the ADC. Maybe a 10 k&Omega; resistor in series and a 5.1 V Zener diode to ground.
I have no idea what current the fuel sender can handle. The power dissipated in it will be VI = 5 &times; 0.166 = 0.83 W.