Why do phasor derivations related to LCR circuits consider $V_c = I * X_c$ even if voltage and current are out of phase? In a capacitor, the current leads by $\pi/2$. But even then derivations of of total voltage in LCR circuits using phasor diagrams assume that $V_{inst.} = I_{inst.} * X_c$ 
Where $V_{inst.}$ is the instantaneous voltage and the $I_{inst.}$ is the instantaneous current and $X_c$ is the capacitive reactance. 
 A: You can't use only capacitive reactance in the equation. You need capacitive impedance. Reactance is the magnitude of capacitive impedance. You also need to use RMS values for voltage and current with phase angles, not instantaneous values, since we are using phasor transformations (frequency domain) of sinusoidal waveforms (time domain).
For ac circuits capacitive impedance can be expressed as either as a complex number (first equation) or phasor (second equation)
$$Z_{C}=-jX_{C}$$
$$Z_{C}=X_{C}\angle -90^0$$
Putting the second in your equation (using phasor voltage) and assuming the phase angle for the voltage is zero, and we get
$$I=\frac {V \angle 0^0}{X_{C}\angle -90^0}$$
$$I= \frac {V\angle+90^0}{X_{C}}$$
And thus current leads voltage by $90^0$.
Hope this helps.
A: 
But even then derivations of of total voltage in LCR circuits using phasor diagrams assume that Vinst. = Iinst. * Xc  

This is not correct.  
The magnitude of the reactance $X_{\rm C}$ is given by $X_{\rm C} = \dfrac {V_{\rm C,peak}}{I_{\rm C,peak}} = \dfrac {V_{\rm C,rms}}{I_{\rm C,rms}}$
