Question

Ce procent de impuritati contine un minereu de siderit, daca din 1500kg minereu s-au obtinut 700kg fier 90% ?

Answer 1

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 2

Answer:

It has 12.84% of impurity

Explanation:

You obtained 700 kg 90% purity of Fe, so you have:

$m_{Fe}=700 kg*0.9=630kg$

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: $FeCO_3$

$M_{siderit}=115.8 kg/kmol$

$M_{Fe}=55.8 kg/kmol$

The mass of siderit:

$m=630 kg Fe*\frac{115.8 kg Siderit}{55.8 kg Fe}=1307.4 kg Siderit$

$m_{impurities}=1500kg-1307.4kg=192.6kg$

Percentaje of impurity:

$P=\frac{192.6kg}{1500kg}*100=12.84$