Question

1) calculate the mass of iron (III) oxide that can be obtained from 480 kg of pure pyrite.
2) what mass of iron could be obtained when the iron oxide obtained from 480kg of pyrite is reduced to metallic iron and oxygen gas?
3) calculate the volume of sulphur dioxide ( measured at rtp) produced from 480g of pyrite.

Answer 1

68681 Chemistry, Other

1) Calculate the mass of iron (III) oxide that can be obtained from 480 kg of pure pyrite.

2) What mass of iron could be obtained when the iron oxide obtained from 480 kg of pyrite is reduced to metallic iron and oxygen gas?

3) Calculate the volume of sulphur dioxide (measured at rtp) produced from 480 g of pyrite.

Answer:

$$4 \mathrm{FeS}_2 + 11 \mathrm{O}_2 = 2 \mathrm{Fe}_2\mathrm{O}_3 + 8 \mathrm{SO}_2$$

1) $n = m/M$

$m = n \cdot M$

M $(\mathrm{FeS}_2) = 120\ \mathrm{g/mol}$

M $(\mathrm{Fe}_2\mathrm{O}_3) = 159.6\ \mathrm{g/mol}$

n $(\mathrm{FeS}_2) = 480000 / 120 = 4000\ \mathrm{mol}$

n $(\mathrm{Fe}_2\mathrm{O}_3) = n(\mathrm{FeS}_2)/2 = 4000/2 = 2000\ \mathrm{mol}$

m $(\mathrm{Fe}_2\mathrm{O}_3) = 2000 \cdot 159.6 = 319200\ \mathrm{g} = 319.2\ \mathrm{kg}$

2) $2\mathrm{Fe}_2\mathrm{O}_3 = 4\mathrm{Fe} + 3\mathrm{O}_2$

n $(\mathrm{Fe}) = 2n(\mathrm{Fe}_2\mathrm{O}_3) = 2000 \cdot 2 = 4000\ \mathrm{mol}$

M $(\mathrm{Fe}_2\mathrm{O}_3) = 159.6\ \mathrm{g/mol}$

M $(\mathrm{Fe}) = 55.8\ \mathrm{g/mol}$

m $(\mathrm{Fe}) = n(\mathrm{Fe}) \cdot M(\mathrm{Fe}) = 4000 \cdot 55.8 = 223200\ \mathrm{g} = 223.2\ \mathrm{kg}$

3) $\mathrm{Fe}_2\mathrm{S}_3 + 9\mathrm{O}_2 = 2\mathrm{Fe}_2\mathrm{O}_3 + 6\mathrm{SO}_2$

n $(\mathrm{SO}_2) = \mathrm{V} / 22.4$

V $(\mathrm{SO}_2) = n \cdot 22.4$

n $(\mathrm{SO}_2) = 6 \cdot n(\mathrm{Fe}_2\mathrm{S}_3) = 6 \cdot 4000 = 24000\ \mathrm{mol}$

V $(\mathrm{SO}_2) = 24000 \cdot 22.4 = 538600\ \mathrm{l} = 538.6\ \mathrm{m}^3$

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