Question

Potassium permanganate, KMnO, and glycerin, C3H5(OH)3, react explosively according to the

following equation:
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
What volume of Co2 at STP can be produced from 701,52 g of KMnO4?
​

Answer 1

The volume of CO2 at STP =124.298 L

Further explanation

Given

Reaction

4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O

701,52 g of KMnO4

Required

volume of CO2 at STP

Solution

mol KMnO4 (MW=158,034 g/mol) :

mol = mass : MW

mol = 701.52 : 158.034

mol = 4.439

mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549

At STP 1 mol = 22.4 L, so for 5.549 moles :

=5.549 x 22.4

=124.298 L