Question

Problema 21 va rog!!!!!! Multumesc mult!!!!

Answer 1

mFe2O3=240 g
%C in cărbune=90 %
mFe=?
VCO2=?
3C+2Fe2O3=4Fe+3CO2
M Fe2O3=2*56+3*16=160 
MCO2=12+2*16=44
3*12g C........2*160g Fe2O3......4*56gFe.......3*44g CO2
x.......................240 g..................z....................y
x=3*12*240/2*160=27 g C
Y= 240*3*44/2*160=99 g CO2

z=240*4*56/2*160=160 g Fe
n Co2=m/M=99/44=2,25 moli
1mol......22,4 l
2,25.........t=50,4lCO2
100g carbune..................90 g C
w.....................................27 g C,w=30 g carbune