# What maximum mass of sulphur in kilograms could be produced?

## Sulphur is used in the production of vulcanized rubber, detergents, dyes, pharmaceuticals, insecticides, and many other substances. The final step in the Claus method for making sulphur is $S {O}_{2} \left(g\right) + 2 {H}_{2} S \left(g\right) \rightarrow 3 S \left(s\right) + 2 {H}_{2} O \left(l\right)$ The first question asked to find the minimum volume of sulphur dioxide at STP that would be necessary to produce 82.5 kg of sulphur. I was able to calculate that $1.92 \cdot {10}^{4} L S {O}_{2}$ was necessary. The second question is the one I am stuck at. What maximum mass of sulphur in kilograms could be produced by the reaction of $\left(1.3 \cdot {10}^{4}\right) {m}^{3}$ of $S {O}_{2} \left(g\right)$ with $\left(2.5 \cdot {10}^{4}\right) {m}^{3}$ of ${H}_{2} S \left(g\right)$, both at STP? Maybe I am just overlooking something obvious, but I feel a bit lost with problem. Thank you very much in advance.

Jul 14, 2018

Question 2

The maximum mass of sulfur that can be produced from the given volumes of $\text{SO"_2}$ and $\text{H"_2"S}$ is $5.5 \times {10}^{4}$ $\text{kg}$.

#### Explanation:

Balanced equation

$\text{SO"_2("g")+ "2H"_2"S(g)}$$\rightarrow$$\text{3S(s) + 2H"_2"O("l")}$

Question 1: Volume of $\text{82.5 kg SO"_2}$

82500color(red)cancel(color(black)("g S"))xx(1"mol S")/(32.06color(red)cancel(color(black)("g S")))="2573 mol S"

2573color(red)cancel(color(black)("mol S"))xx(1"mol SO"_2)/(3color(red)cancel(color(black)("mol S")))="857.7 mol SO"_2"

At STP of ${0}^{\circ} \text{C}$ and $\text{1 atm}$, the molar volume of a gas is $\text{22.414 L/mol}$. To calculate the volume of $\text{857.7 mol SO"_2}$, multiply mol $\text{SO"_2}$ by the molar volume of a gas.

857.7color(red)cancel(color(black)("mol SO"_2))xx("22.414 L")/(1color(red)cancel(color(black)("mol")))=1.92xx10^4 $\text{L SO"_2}$

Question 2: Maximum mass of sulfur in kilograms

Part 1) $1.3 \times {10}^{4}$ $\text{m"^3}$ $\text{SO"_2}$

Convert $\text{m"^3}$ to $\text{L}$.

$\text{1 m"^3}$$=$$\text{1000 L}$

1.3xx10^4color(red)cancel(color(black)("m"^3))"SO"_2xx("1000 L")/(1color(red)cancel(color(black)("m"^3)))=1.3xx10^7 $\text{L SO"_2}$

We can determine moles $\text{SO"_2}$ using the molar volume. Multiply the volume in liters by the inverse of the molar volume.

1.3xx10^7color(red)cancel(color(black)("L")) "SO"_2xx(1"mol")/(22.414color(red)cancel(color(black)("L")))=5.8xx10^5 $\text{mol SO"_2}$

Determine moles $\text{S}$ by multiplying mol $\text{SO"_2}$ by the mole ratio between $\text{S}$ and $\text{SO"_2}$ in the balanced equation, with $\text{S}$ in the numerator.

5.8xx10^5color(red)cancel(color(black)("mol SO"_2))xx(3"mol S")/(1color(red)cancel(color(black)("mol SO"_2")))=1.7xx10^6 $\text{mol S}$

Determine the mass of $\text{S}$ by multiplying mol $\text{S}$ by its molar mass $\left(\text{32.06 g/mol}\right)$.

1.7xx10^6color(red)cancel(color(black)("mol S"))xx(32.06"g S")/(1color(red)cancel(color(black)("mol S")))=5.5xx10^7 $\text{g S}$

Convert mass in grams to kilograms.

5.5xx10^7color(red)cancel(color(black)("g")) "S"xx("1 kg")/(1000color(red)cancel(color(black)("g")))=5.5xx10^4 $\text{kg S}$

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Part 2) $2.5 \times {10}^{4}$ $\text{m"^3" H"_2"S}$

Convert $\text{m"^3}$ to $\text{L}$.

2.5xx10^4color(red)cancel(color(black)("m"^3)) "H"_2"S"xx("1000 L")/(1color(red)cancel(color(black)("m"^3")))=2.5xx10^7 $\text{L H"_2"S}$

To get the moles of $\text{H"_2"S}$, multiply the volume in liters by the inverse of the molar volume.

2.5xx10^7color(red)cancel(color(black)("L")) "H"_2"S"xx(1"mol")/(22.414color(red)cancel(color(black)("L")))=1.1xx10^6 $\text{mol H"_2"S}$

Determine moles $\text{S}$ by multiplying mol $\text{H"_2"S}$ by the mole ratio between $\text{S}$ and $\text{H"_2"S}$ in the balanced equation, with $\text{S}$ in the numerator.

1.1xx10^6color(red)cancel(color(black)("mol H"_2"S"))xx(3"mol S")/(2color(red)cancel(color(black)("mol H"_2"S")))=1.7xx10^6 $\text{mol S}$

Determine the mass of $\text{S}$ by multiplying mol $\text{S}$ by its molar mass $\left(\text{32.06 g/mol}\right)$.

1.7xx10^6color(red)cancel(color(black)("mol S"))xx(32.06"g S")/(1color(red)cancel(color(black)("mol S")))=5.5xx10^7 $\text{g S}$

Convert mass in grams to kilograms.

5.5xx10^7color(red)cancel(color(black)("g")) "S"xx(1"kg")/(1000color(red)cancel(color(black)("g")))=5.5xx10^4 $\text{kg S}$

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The maximum mass of sulfur that can be produced from the given volumes of $\text{SO"_2}$ and $\text{H"_2"S}$ is $5.5 \times {10}^{4}$ $\text{kg}$.