Answer to Question #99540 in General Chemistry for Beverlie

Question #99540
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?
1
Expert's answer
2019-12-03T08:24:20-0500

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\\%."


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