A.)Calculate the Mass of manganese IV oxide that can be synthesized from 15.00 grams of potassium iodide.
B) calculate the percent yield of this experiment if mass of 1.982 grams of manganese IV oxide is produced.
C) determine the percent error.
D) compute for the excess amount of the excess reactant.
A). 6KI(aq) +4H2O(l)+2KMnO4(aq)"\\to" 3l2(s) +2 MnO2(s)+8KOH(aq)
REM of KI=166.00 g/mol
Moles of KI=15/ 166
= 0.09036moles
Mole ratio= KI:MnO2 = 6:2 = 3:1
Moles of MnO2= 0.09036/3
=0.03012moles
REM of MnO2= 86.99 g/mol
mass of MnO2= 0.03012× 86.97
=2.6195364g
B)Percent Yield =44.47%
C)Percent error is determined by the difference between the exact value and the approximate value of a quantity, divided by the exact value and then multiplied by 100 to represent it as a percentage of the exact value.
D)To find the amount of remaining excess reactant, subtract the mass of excess reagent consumed from the total mass of excess reagent given.
Comments
Leave a comment