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3 years ago

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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3 years ago

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