Question

47.25g of MnO2 is reacted with 48.9g of HCl in the following reaction.
MnO2 + HCl=MnCl2+Cl2+ H20
a. what is the limiting reagent and calulate the theorerical yeield of Cl2
b. how much of the excess reagent is remaining
c if the found actual yeild is 19.65g what is the percent yeild?

Answer 1

Lets first equalize the equation:

$MnO_2+4HCl \rightarrow MnCl_2+Cl_2 \uparrow +2H_2O;$

a) The amount of the reactants in moles is:

$n(MnO_2)=\frac{m(MnO_2)}{M(MnO_2)}=\frac{47.25g}{86.94g/mol}=0.544mol;$

$n(HCl)=\frac{m(HCl)}{M(HCl)}=\frac{48.9g}{36.46g/mol}=1.34mol;$

According to the equation, the reactants must be taken in the 1 to 4 ratio. Thus, HCl is a limiting reactant.

The amount of Cl₂ in moles is 0.335 mol (see equation) and the mass of Cl₂:

$m(Cl_2)=0.335mol*70.91g/mol=23.8g;$

The theor. yield of Cl₂ is 23.8 g.

b) If HCl is a limiting reagent, then MnO₂ is an excess reagent. Lets calculate the amount of MnO2 reacted:

$n(MnO_2)_{react}=0.335mol;$

$n(MnO_2)_{ex}=0.544mol-0.335mol=0.209mol;$

$m(MnO_2)_{ex}=0.209mol*86.94g/mol=18.17g;$

c) Lets calculate the practical yield of chlorine:

$\gamma = \frac{m_{act.}}{m_{theor.}}*100\%=\frac{19.65g}{23.8g}*100\%=82.6\%.$