stehiometrija

Submitted by lejla93 on Mon, 03/24/2014 - 20:56
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koliko tona sumpora  mora biti spaljeno da bismo dobili 50 000 m(3) SO2 pri 750mmHg i 80 stepeni C?

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milicaradenkovic - Mon, 03/24/2014 - 21:27

S+O2------------------->SO2

n(SO2)=V(SO2)*p/R*T

p=750*133.322=99.99 kPa

n(SO2)=50000*99.99*10(3)/8.314*353.15

n(SO2)=1.703*10(6) mol

n(S)=n(SO2)=1.703*10(6) mol

m(S)=n(S)*Mr=1.703*10(6)*32=54.49 t