Question

Mg(MnO4)2 + K2Cr2O7 -> MgCr2O7 + 2 KMnO4

11.2 grams of magnesium permanganate reacts with 14.9 grams of potassium dichromate to produce 13.8 grams of magnesium dichromate and potassium permanganate.

What substance is the limiting reagent? Show all work.

What is the theoretical yield of magnesium dichromate? Show all work.

What is the percent yield? Show all work.

Answer 1

Atomic weights : Mg= 24, Mn=55, O=16,Cr=52, K=39

molar mass ( gm/mole): Mg(MnO4)2= 24+2{55+64)=262 , K2Cr2O7= 2*39+2*52+7*16=294, MgCr2O7= 24+2*52+7*16=240 and KMnO4= 39+55+4*16=158

moles= mass/molar mass, moles : Mg(MnO4)2= 11.2/262=0.043, K2Cr2O7=14.9/294=0.051

molar ratio of Mg(MnO4)2 : K2Cr2O7 ( actual )= 0.043 :0.051= 1: 0.051/0.043= 1:1.19

Theoretical ratio as per the given reaction= 1:1

hence K2Cr2O7 is excess reactant. Limiting reactant is Mg(MnO4)2

1 mole of Mg(MnO4)2 gives 1 mole of MgCr2O7

0.043 moles of Mg(MnO4)2 gives 0.043 moles of MgCr2O7 and 0.086 moles of KmnO4, mass of MgCr2O7= moles* Molar mass= 0.043*240 =10.32 gm, mass of KMnO4= 0.086*158=13.6 gm

mass ratio of MgCr2O7: KMnO4= 10.32 : 13.6

hence mass of MgCr2O7 in 13.8 gm of products produced = 13.6*10.32/ (10.32+13.6)=5.9 gm

% yield =100*5.9/10.32= 57.2%