Answer to Question #256973 in General Chemistry for Jonalyn

Question

Kl(aq) + H2O(l) + KMnO4(aq) 166.0/mol 18.02g/mol 153.03g/mol l2(s) + MnO2(s) + KOH(aq) 253.80g/mol 86.97g/mol 56.11g/mol A.) Calculate the mass of manganese (lV) oxide that can be synthesized from 15.00 grams of potassium iodide. B.) Calculate the percent yield for this experiment if a mass of 1.982 grams of manganese (lV) oxide is produced. C.) Determine the percent error D.) compute for the excess amount of the excess reactant

Answer 1

6KI_(aq) +4H₂O_(l)+2KMnO_4(aq)$\to$ 3l_2(s) +2 MnO_2(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

Percent Yield =44.47%