Refer to the above block diagram. If the flow rate of stream $1=180k\frac{g}{m}in,$ then:

The mass flow rate of solubles in stream $1=k\frac{g}{m}in(2$ s$.$f$)$

The mass flow rate of insolubles in stream $1=k\frac{g}{m}in(3$ s$.$f$)$

The mass flow rate of insolubles in stream $6=k\frac{g}{m}in(3s.f)$

The total mass flow rate of stream $6=k\frac{g}{m}in(3s.f)$

Adapted from Q$.3\mathrm{.}14$ of T$.$ Michael Duncan & Jeffrey A$.$ Reimer, $"$ Chemical Engineering Design and Analysis $-$ An Introduction", Cambridge $1998$