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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, $$\begin{array}{rl}r(x)& =6y^3+5y^2-16y+8-(2y+3)(3y^2+cy+\frac{5}{2}c)\\ & =6y^3+5y^2-16y+8-(6y^3+2c{y}^2+5cy+9y^2+3cy+\frac{15}{2}c)\\ & =(5-9-2c)y^2-(16+5c+3c)y+8-\frac{15}{2}c\\ & =(-4-2c)y^2-(16+8c)y+8-\frac{15}{2}c\end{array}$$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 $$r(x)=(-4+4)y^2-(16-16)y+8-\frac{15}{2}\cdot (-2)=8+15=\boxed{23}.$$