If the remainder is $ r(x) $, we know that $$6y^3+5y^2-16y+8=(2y+3)(3y^2+cy+\frac{5}{2}c) + r(x).$$So, $$\begin{aligned} r(x) &= 6y^3+5y^2-16y+8 - (2y+3)(3y^2+cy+\frac{5}{2}c)\\ &=6y^3+5y^2-16y+8-(6y^3+2cy^2+5cy+9y^2+3cy+\frac{15}{2}c) \\ &=(5-9-2c)y^2-(16+5c+3c)y+8-\frac{15}{2}c \\ &=(-4-2c)y^2-(16+8c)y+8-\frac{15}{2}c \\ \end{aligned}$$Since the divisor $ 2y+3 $ is linear we know that the remainder must be constant. So, $$-4-2c = 0$$which means $$c = -2.$$So the remainder is $$r(x) = (-4+4)y^2-(16-16)y+8-\frac{15}{2}\cdot(-2) =8+ 15 = \boxed{23}.$$