Molar balance equation example problems

In previous lesson, we have obtained molar balance equations for different reactors. Here we are going to apply that molar balance equation for a CSTR reactor to get familiar with that equation.

CSTR reactor

CSTR means continuous stirred tank reactor


There is a CSTR which has inflow and outflow operates at steady state. A and B are reactants which enters to the reactor at a rate of 10.01 mol/min. A and B is react and produce C and D.
A + B → 2C + D.
at steady state A is reacted at a rate of 8.94 mol/min
Calculate followings.
a) producing rate of products
b) rate of A and B in inflow

general reactor flow rates



According to the reaction, one A molecule and one B molecule react together and produce two C molecules and one D molecule. Therfore producing rate of C is twice as reducing rate of A.

producing rate of C = 8.94 mol/min * 2 = 17.88 mol/min
producing rate of D = 8.94 mol/min


molar balace short equation

This reaction occurs in steady state. Therefore no accumulation in this reactor. Here we are going to apply our quantities in moles. Our time basis is per minute.

Apply molar balance for A

A is reacting and reducing in the reactor. So generation is negative.

flow rate of A in inflow, FA|in = 10.01 mol/min

generation of A, rA= -8.94 mol/min

accumulation = 0

substitute those values to the molar balance equation,

10.01 + ( - 8.94 ) = FA|out + 0

FA|out = 1.07 mol/min

Flow rates of A and B in inflow is same. Also A and B reacts 1:1. therefore reaction rate of A and B is same and also flow rate of B in outflow is same as flow rate of A in outflow.
FB|out = 1.07 mol/min

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