Question

One of the products formed as a result of degradation reaction by heating KClO3 is O2 (g). How many liters is the volume of O2 (g) formed by the degradation of 0,018 mol KClO3 at 750mmHg and 200C?

Answer 1

The reaction is $2KClO_3 \to 2KCl+3O_2$

According to the reaction equation, find the chemical amount of oxygen: 0,018mol*3/2=0,027mol

Using the Clapeyron-Mendeleev equation: $PV=nRT$ , P - gas pressure (Pa) , V - gas volume (m³), n - chemical amount of gas (mol), T - temperature of gas (K), R - Gas constant (8,314 J/(mol*K)).

P=750mmHg=99991,8Pa

T=200oC=473K

Express volume: $V=nRT/P$ ; V=(0,027mol*8,314J/(mol*K)*473K)/99991,8Pa=0,0011m³

V=0,0011m³=1,1L

Answer:1,1L