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23 April, 12:18

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

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  1. 23 April, 12:25
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    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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