Question

Problema chimie clasa a 8 a:
400 g oxid de fier (FE2O3) de puritate 80% reactioneaza cu carbune de puritate 90%.
Se cere:
a) formula
b) masa de fier
c) nr de molecule de CO2 rezultă...
d) masa de carbune folosit​

Answer 1

Răspuns:

2Fe2O3+3C->4Fe+3CO2

p=mp*100/mip

mp-masa pura

mip-masa impura

p-puritate

mp=p*mip/100=320g pur

c)

320g Fe2O3-------------------------xgCO2=132g

2*160............................................44*3

niu=m/M=3 moli

N=niu*Na=18,069*10^23 molecule

d)

320g FE203--------------------------YgC=36g

2*160............................................3*12

y=m impura

m pura=p*mip/100=32,4g pur

...............................................................................................................................

Coroana?

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