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sattari [20]
3 years ago
10

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

Chemistry
1 answer:
Pachacha [2.7K]3 years ago
8 0

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




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