P$1-5{?}_A$ OEQ $($Old Exam Question$)$ The reaction

$A+B2C$

takes place in an unsteady CSTR$.$ The feed is only A and $B$ in equimolar proportions. Which of the following sets of equations gives the correct set of mole balances on A$,$

$B,$ and $C?$ Species A and B are disappearing and species C is being formed. Circle the correct answer where all the mole balances are correct. $($a$)($b$)$

$F_{B0}-F_A-{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_A}{dt}$

$F_{B0}-F_B-{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_B}{dt}$

$-F_C+2{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_C}{dt}$

$F_{A0}-F_A+{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_A}{dt}$

$F_{A0}-F_B+{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_B}{dt}$

$-F_C-2{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_C}{dt}$

$F_{A0}-F_A+{\displaystyle \underset{0}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_A}{dt}$

$F_{A0}-F_B+{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_B}{dt}$

$-F_C+{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{C}dV=\frac{d{N}_C}{dt}$

$F_{B0}-F_A-{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_A}{dt}$

$F_{B0}-F_{A0}-{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{A}dV=\frac{d{N}_B}{dt}$

$-F_C+{\displaystyle \underset{\phantom{}}{\overset{V}{}}}{r}_{C}dV=\frac{d{N}_C}{dt}$