How to calculate MOSFETs gate resistor?
I don't want to make this a one-link answer, so I'll write a quick summary, but you should really read this appnote.


  If I use the equation Q=CV I can find the charge


Not really. MOSFET capacitance varies with Vgs and Vds. Also, a large part of the gate charge is due to miller effect through Cgd:



First the FET is OFF, and Vds is usually equal to the supply voltage Vcc. Then, gate voltage rises as current charges Cgs. Once threshold voltage is reached, the FET begins to conduct, and Vds goes down. This causes the voltage across Cgd to vary, and a capacitor with varying voltage across it implies a current. Thus the driver has to inject all the charge required to bring the voltage across Cgd to its final value. Once the FET is fully ON, Vds is quite small, and then gate current is once again used to ramp up Vgs and reduce RdsON.

During switching, Cgd varies a lot depending on Vds, so you cannot use Q=CV which implies a constant capacitor. You must use datasheet values instead, or simulation with accurate models.

Total gate charge will depend on final Vgs, but also on initial Vds (=supply voltage). Your calculation ignores Cgd, so you are under estimating Qg.

Now your original question about the gate resistor value. It's a bit complicated. Higher resistor value will slow down turn-on but calculating how much time it will take to charge the gate up to Qg is always a big approximation as the MOSFET gate is not a constant value capacitor, and the MOSFET driver usually outputs a voltage, so using a fixed value resistor will result in high current during the beginning of turn-on, but as gate voltage increases then current will get lower... like in a RC circuit.

The purpose of the gate resistor is to prevent MOSFET oscillation, slow down switching if you want to avoid EMI problems, stuff like that. If you use a low frequency then slowing down the switching is an excellent way to reduce EMI. Some circuits use different resistors for turn-on and turn-off, with diodes or a dual-output driver ; this is a way to adjust switching times to avoid cross-conduction when using two FETs in synchronous mode. 

So the value of the resistor depends on what its intended use is... which you don't tell.