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Elan Coil [88]
3 years ago
12

Se realiza una mezcla de minerales de Cu y Fe: 20 kg FeS2 (pirita), 70 kg de Fe2O3 (hemetita) 15 kg de CuFe2 (calcopirita) y 90

kg de CuO (tenorita). Calcule: Porcentaje de hierro Porcentaje de cu Porcentaje de azufre y oxigeno
Chemistry
1 answer:
Artemon [7]3 years ago
8 0

Answer:

34.78% Fe

39.66% Cu

5.48% S

20.07% O

Explanation:

Para resolver esta pregunta debemos hallar la masa de cada átomo en cada mineral. Así, podremos hallar el porcentaje de cada átomo:

<em></em>

<em>Pirita (Fe: 55.845g/mol; S: 32.065g/mol; FeS2: 119.975g/mol)</em>

Masa Fe:

20kg FeS2 * (1*55.845g/mol / 119.975g/mol) = <em>9.31kg Fe</em>

Masa S:

20kg FeS2 * (2*32.065g/mol / 119.975g/mol) = <em>10.69kg S</em>

<em>Hemetita (Fe: 55.845g/mol; O: 16g/mol; Fe2O3: 159.688g/mol)</em>

Masa Fe:

70kg Fe2O3 * (2*55.845g/mol / 159.688g/mol) = <em>48.96kg Fe</em>

Masa O:

70kg Fe2O3 * (3*16g/mol / 159.688g/mol) = <em>21.04kg O</em>

<em>Calcopirita (Fe: 55.845g/mol; Cu: 63.546g/mol; CuFe2: 175.236 g/mol)</em>

Masa Fe:

15kg CuFe2 * (2*55.845g/mol / 175.236 g/mol) = <em>9.56kg Fe</em>

Masa Cu:

15kg CuFe2 * (1*63.546g/mol / 175.236 g/mol) = <em>5.44kg Cu</em>

<em />

<em>Tenorita (O: 16g/mol; Cu: 63.546g/mol; CuO: 79.545 g/mol)</em>

Masa O:

90kg CuO * (1*16g/mol / 79.545 g/mol) = <em>18.10kg O</em>

Masa Cu:

90kg CuO * (1*63.546g/mol / 79.545 g/mol) = <em>71.90kg Cu</em>

<em />

Masa Total: 20kg + 70kg + 15kg + 90kg = 195kg

Porcentaje Hierro:

9.31kg Fe + 48.96kg Fe + 9.56kg Fe / 195kg * 100 =

34.78% Fe

Porcentaje Cobre:

5.44kg Cu + 71.90kg Cu / 195kg * 100 =

39.66% Cu

Porcentaje Azufre:

10.69kg S / 195kg * 100 =

5.48% S

Porcentaje Oxígeno:

21.04kg O + 18.10kg O/ 195kg * 100 =

20.07% O

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