# A compound containing "36.5%" of sulfur and "63.5%" of iron reacts with a compound containing "27.6%" oxygen and "724%" of iron. What is the balanced equation and the theoretical yield of iron oxide?

##### 2 Answers
Jun 18, 2015

The theoretical yield of the iron oxide will be 880 g.

#### Explanation:

The first thing you have to do is determine the formula for iron sulfide by using the known percent composition of sulfur in the compound. Let's assume that the formula for iron sulfide looks like this

$F {e}_{x} {S}_{y}$

The percent composition of sulfur in this formula would be

$\left(y \cdot 32.07 \cancel{\text{g/mol"))/((x * 55.85 + y * 32.07)cancel("g/mol}}\right) \cdot 100 = 36.5$

This means that you have

$y \cdot 32.07 \cdot 100 = x \cdot 55.85 \cdot 36.5 + y \cdot 32.07 \cdot 36.5$

$3207 \cdot y = 20.38 .525 \cdot x + 1170.555 \cdot y$

$2036.445 \cdot y = 2038.555 \cdot x \implies \frac{x}{y} = \frac{2036.445}{2038.555} = 0.9990 \cong 1$

The ratio between iron ans dulfur in rion sulfide is $1 : 1$. This means that the chemical formula of the compound will have to be $F e S$.

The chemical equation looks like this

$F e S + {O}_{2} \to F {e}_{a} {O}_{b} + S {O}_{2}$

Now do the same for the iron oxide.

$\left(b \cdot 16.0 \cancel{\text{g/mol"))/((a * 55.85 + b * 16.0)cancel("g/mol}}\right) \cdot 100 = 27.6$

This will get you

$1600 \cdot b = 1541.46 \cdot a + 441.6 \cdot b$

$1158.4 \cdot b = 1541.46 \cdot a \implies \frac{a}{b} = \frac{1158.4}{1541.46} = 0.751 \cong \frac{3}{4}$

If you take $b$ to be $\frac{4}{3} \cdot a$, your oxide will look like this

$F {e}_{a} {O}_{\left(\frac{4}{3}\right) \cdot a} = F {e}_{1} {O}_{\frac{4}{3}}$ $\to$ multiply by 3 to get $F {e}_{3} {O}_{4}$.

This means that your balanced chemical equation will look like this

$\textcolor{red}{3} F e {S}_{\left(s\right)} + 5 {O}_{2 \left(g\right)} \to F {e}_{3} {O}_{4 \left(s\right)} + 3 S {O}_{2 \left(g\right)}$

Notice the $\textcolor{red}{3} : 1$ mole ratio that exists between iron sulfide and iron (II, III) oxide.. This tells you that, regardless of how many moles of iron sulfide react, your reaction will produce 5/3 fewer moles of the iron oxide.

Since oxygen is in excess, all the moles of iron sulfide will take part in the reaction. Use the compound's molar mass to determine how many moles you have in your 1.0-kg sample

1.0cancel("kg") * (1000cancel("g"))/(1cancel("kg")) * ("1 mole "FeS)/(87.91cancel("g")) = "11.38 moles" $F e S$

This means that the reaction will produce

11.38cancel("moles"FeS) * ("1 mole "Fe_3O_4)/(color(red)(3)cancel("moles"FeS)) = "3.793 moles" $F {e}_{3} {O}_{4}$

The theoretical yield of the reaction will thus be

3.793cancel("moles"Fe_3O_4) * "231.53 g"/(1cancel("mole"Fe_3O_4)) = "878.19 g" $F {e}_{3} {O}_{4}$

Rounded to two sig figs, the number of sig figs you gave for the mass of iron sulfide, the answer will be

${m}_{F {e}_{3} {O}_{4}} = \textcolor{g r e e n}{\text{880 g = 0.88 kg}}$

Jun 18, 2015

The balanced equation is ${\text{3FeS"+"5O}}_{2}$$\rightarrow$$\text{Fe"_3"O"_4"+3SO"_2}$.

The theoretical yield of ${\text{Fe"_3"O}}_{4}$ is 880 g. (Rounded to two significant figures due to 1.0 kg.)

#### Explanation:

A) BALANCED EQUATION

In order to write a balanced chemical equation for this reaction, we must determine the formulas for iron sulfide and iron oxide, as they have more than one formula because iron forms more than one ion.

Determine the formulas for iron sulfide and iron oxide.

Iron Sulfide

S=36.5%
Fe=63.5%

Because 36.5% is 35/100, we can assume a 100-g sample of iron sulfide, and we can say that there are 36.5 g S and 63.5 g Fe. We can convert the masses to moles by dividing the mass of each element by its molar mass (atomic mass on the periodic table in grams.

Sulfur

(36.5 "g S")/((32.065 "g S")/("1 mol S"))=((36.5 cancel"g S")xx(1 "mol S")/(32.065 cancel"g S"))=1.138 "mol S"

Iron

(63.5 "g Fe")/((55.845 "g Fe")/(1 "mol S"))=((63.5 cancel"g Fe")xx(1 "mol Fe")/(55.845 cancel"g Fe"))=1.137 "mol Fe"

Mole ratios for $\text{Fe}$ and $\text{S}$

$\left(1.138 \cancel{\text{mol")/(1.137 cancel"mol}}\right) = 1.000$

Iron and sulfur are present in a 1:1 ratio.

The formula for iron sulfide is $\text{FeS}$.

Iron Oxide

Again, because 27.6% is 27.6/100, we assume a 100-g sample of iron oxide, and we can say that there are 27.6 g oxygen and 72.4 g Fe. Again, we can convert the masses to moles by dividing the mass of each element by its molar mass (atomic mass on the periodic table in grams.

Oxygen

(27.6 "g O")/((15.999 "g O")/(1 "mol O"))=((27.6 "g O")xx(1"mol O")/(15.999 "g O"))=1.725 "mol O"

Iron

(72.4 "g Fe")/((55.845 "g Fe")/(1"mol Fe"))=((72.4 cancel"g Fe")xx(1"mol Fe")/(55.845 cancel"g Fe"))=1.296 "mol O"

Mole ratios for O and Fe

Divide the moles of each element by the smallest number of moles.

$\text{O} :$$=$$\left(1.725 \text{mol")/(1.296"mol}\right) = 1.331$

$\text{Fe} :$$=$$\left(1.296 \text{mol")/(1.296"mol}\right) = 1.000$

Mole ratios must be in whole numbers, so multiply both ratios times $3$.

$\text{O} :$$=$$\left(1.725 \text{mol")/(1.296"mol}\right) = 1.331 \times 3 = 3.993 \approx 4$

$\text{Fe} :$$=$$\left(1.296 \text{mol")/(1.296"mol}\right) = 1.000 \times 3 = 3$

The mole ratio for $\text{Fe:O}$$=$$3 : 4$

The chemical formula for iron oxide is $\text{Fe"_3"O"_4}$.

Balanced Chemical Equation

$\text{3FeS"+"5O"_2}$$\rightarrow$$\text{3SO"_2"+Fe"_3"O"_4}$

B) THEORETICAL YIELD OF IRON OXIDE

We need the molar masses of $\text{FeS}$ and $\text{Fe"_3"O"_4}$ and the mole ratio from the balanced equation.

Molar mass of $\text{FeS"=(1xx55.845 "g/mol")+(1xx32.065 "g/mol")=87.910 "g/mol}$

Molar mass of "Fe"_3"O"_4=(3xx55.945 "g/mol")+(4xx15.999 "g/mol")=231.531 "g/mol"

Mole ratio of iron oxide and iron sulfide is:

$\left(1 \text{mol" "Fe"_3"O"_4)/(3 "mol FeS}\right)$

Calculate Theoretical Yield of Iron Oxide

Convert 1 kg FeS to 1000 g FeS. Then convert the mass of FeS to moles using its molar mass. Multiply moles FeS times the mole ratio between iron oxide and iron sulfide to get moles iron oxide. Multiply moles of iron oxide times its molar mass.

$1000 \cancel{\text{g FeS"xx(1cancel"mol FeS")/(87.91 cancel"g FeS")xx(1cancel"mol Fe"_3"O"_4)/(3cancel"mol FeS")xx(231.531 "g Fe"_3"O"_4)/(1cancel"mol Fe"_3"O"4)=880 "g Fe"_3"O"_4}}$
(answer rounded to two significant figures because of 1.0 kg)