FeCO3 ==> Fe + produsi minoritari.
m Fe impur= 700 kg
puritatea (p) = masa pura (mp)/ masa impura (mi) x 100
mp= p x mi / 100 sau mp = p/100 x mi => mp Fe = 90/100 x 700 = 630 kg Fe pur.
M FeCO3= 115.85 kg/kmol
115.85 kg FeCO3 .... 55.85 kg Fe
x kg FeCO3 ........630 kg Fe
x= 630 * 115.85 /55.85 = 1306.81 kg FeCO3 (mp in formula puritatii)
p=mp/mi x 100
mi FeCO3 = 1500 kg
mp FeCO3=1306.81 kg
p=1306.81 / 1500 x 100 = 87.12% puritate Siderit
Answer:
It has 12.84% of impurity
Explanation:
You obtained 700 kg 90% purity of Fe, so you have:
[tex]m_{Fe}=700 kg*0.9=630kg[/tex]
Those 630 kg are the mass of Fe contained in the original sample of 1500 kg in the from of Siderit.
The other substeances that aren't Siderit in the sample can be considered impurities.
Siderit: [tex]FeCO_3[/tex]
[tex]M_{siderit}=115.8 kg/kmol[/tex]
[tex]M_{Fe}=55.8 kg/kmol[/tex]
The mass of siderit:
[tex]m=630 kg Fe*\frac{115.8 kg Siderit}{55.8 kg Fe}=1307.4 kg Siderit[/tex]
[tex]m_{impurities}=1500kg-1307.4kg=192.6kg[/tex]
Percentaje of impurity:
[tex]P=\frac{192.6kg}{1500kg}*100=12.84[/tex]
