Question

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 a 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.

Answer 1

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

A) = 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.