in the reaction below 25.00g of MnCl2 is reacted with 100.0g of PbO2, excess KCI
ID: 840303 • Letter: I
Question
in the reaction below 25.00g of MnCl2 is reacted with 100.0g of PbO2, excess KCI and excess HCI
2KCI(aq) + 2MnCI2 (aq) +5PbO2(s)+4HCI(aq)------> 2KMnO4(aq) +5PbCI2(s)+2H2O(L)
a) how many grams of KMnO can be produced by this reaction? Use an ICE table.
b) if 21.42 g of KMnO4 is actually produced, what is the percent yield?
c) how many grams of what starting substances will be left over after the reaction ?
d) How many grams of what starting substance (i.e MnCl2 or PbO2) must be added to the original quantities of reactants so that there will be neither PbO2 nor MnCl2 left over after the reaction ?
Explanation / Answer
B)
moles MnCl2 = 15.00/125.84 g/mol=0.199
moles PbO2 required = 0.199 x 5 /2=0.498
moles PbO2 = 100.0 g/ 239.198 g/mol= 0.418 so PbO2 is the limiting reactant
theoretical moles KMnO4 = 0.418 x 2/5=0.167
theoretical mass KMnO4 = 0.167 x 159.034 g/mol=26.39
% yield = 21.42 x 100/ 26.39=81.16
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