Question

Iron pyrite (FeS2) is the form in which much of the sulfur exists in coal. In the combustion of coal, oxygen reacts with iron pyrite to produce iron(III) oxide and sulfur dioxide, which is a major source of air pollution and a substantial contributor to acid rain. What mass of Fe2O3 is produced from 84 L of oxygen at 3.41 atm and 148◦C with an excess of iron pyrite? Answer in units of g.

Answer 1

4FeS₂ + 11O₂ = 2Fe₂O₃ + 8SO₂↑

n(FeS₂) = m(FeS₂)/M(FeS₂) = 6000/120 = 50 моль

по уравнению реакции - n(O₂) = (11*n(FeS₂))/4 = (11*50)/4 = 137,5 моль

V(O₂) = n(O₂)*Vm = 137,5*22,4 = 3080 л

ответ: V(O₂) = 3080 л

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Answer 2

Answer:

$8.297 mol O_2 *\frac{2 mol Fe_2 O_3}{11 mol O_2} *\frac{159.69 gr Fe_2 O_2}{ 1mol Fe_2 O_3}= 240.90 gr Fe_2 O_3$

Explanation:

For this case we have the following equation:

$4FeS_2 +11 O_2 \rightarrow 2Fe_2 O_3 +8SO_2$

After balance the equation.

For this case we can use the following expression for an ideal gas:

$PV = nRT$

And if we solve for $n_{O_2}$ we got:

$n_{O_2} = \frac{PV}{RT} =\frac{3.41 atm *84L}{ 0.082 \frac{atm L}{mol K} (148+273)K} =8.297 mol O_2$

The mass for the $Fe_2 O_3$ is given by: $55.845gr*2 +16gr*3=159.69 gr$, since we have 2 molecules of iron and 3 from oxygen.

For this case since the pyrite is the excess reactive we can calculate the amount of $Fe_2 O_3$ with the oxygen and we can do this:

$8.297 mol O_2 *\frac{2 mol Fe_2 O_3}{11 mol O_2} *\frac{159.69 gr Fe_2 O_2}{ 1mol Fe_2 O_3}= 240.90 gr Fe_2 O_3$