# Chemistry: The Central Science 14th edition, chapter 3, Design an experiment problem

I am new to chemistry as well as StackExchange network, so any advice is appreciated.

I tried to answer it myself, but I am sure it is not completely correct and there are certainly alternative solutions. There are also points I would like to know which I listed at the end of my solution.

Here's what I came up with: a) We assume $$S+O_2 \rightarrow SO_2$$. The equation is balanced and $$1 \space mol \space S≏ 1 \space mol \space O_2 ≏ 1 \space mol \space SO_2$$ thus, $$0.1 \space mol$$ $$O_2$$ is needed for $$0.1 \space mol$$ $$S$$ to react.

b) We assume $$S+O_2 \rightarrow SO_3$$. Balancing the equation gives $$2S+3O_2 \rightarrow 2SO_3$$ and $$2 \space mol \space S≏ 3 \space mol \space O_2 ≏ 2 \space mol \space SO_2$$ thus, $$0.15 \space mols$$ $$O_2$$ is needed for $$0.1 \space mol$$ $$S$$ to react.

c) Equipment given is an analytical balance, which have an accuracy of $$0.0001$$ to $$0.00001$$ grams. Since we know the molar mass of Oxygen (gas) is $$31.999$$ g and that of Sulfur is $$32.065$$ grams, we can weigh them on the balance to use correct number of moles. (For practical purposes, we can approximate both of these quantities as $$32$$ grams)

d) At room temperature, Sulfur Dioxide is a colorless gas meanwhile Sulfur Trioxide is a crystalline solid. Their physical properties can be used to identify them. I am unsure about the instruments.

e) I would use $$1 \space mol \space S$$ and $$1 \space mol \space O_2$$. If the assumption a) is true, then the reaction will result in one mole of Sulfur Dioxide. If no solid is left on the reaction vessel then assumption a) is supported. If the assumption b) is true, then $$O_2$$ will be the limiting reagent and products of the reaction will be $$\frac{2}{3} \space mol$$ Sulfur Trioxide and $$\frac{1}{3} \space mol$$ leftover Sulfur. Sulfur is also a solid in room temperature and if these two solids weigh 64 grams (that is, the weight of reactants) then assumption b) is supported. The last case is when the product is a mixture of $$SO_2$$ and $$SO_3$$, let's call it assumption c). Balanced equation then becomes: $$4S+5O_2 \rightarrow 2SO_2+2SO_3$$. Since Oxygen is still the limiting reagent, we expect products of the reaction to be $$\frac{2}{5} \space mol \space SO_2$$, $$\frac{2}{5} \space mol \space SO_3$$ and $$\frac{1}{5} \space mol \space S$$.

Points I would like to know:

• What are the ways to distinguish between Sulfur Dioxide gas and Oxygen gas besides the impractical (downright foolish) idea of inhaling the resulting gas and checking for poisoning symptoms?
• What could I have used the furnace for?
• If I carry out the experiment, will I get Sulfur Dioxide or can initial conditions like temperature affect the results?
• You cannot use $1$ mole $\ce{O2}$, because $1$ mol of any gas has a volume of about $25$ liters at room temperature and pressure. No laboratory vessel has a volume greater than $1$ liter. The furnace is here to heat the sulfur. Sulfur does not burn at room temperature. It has to be heated to about $200$°C to start burning. The amount of oxygen will be easier determined by knowing the volume of the vessel, and deriving the number of moles by $\pu{n = pV/RT}$ May 26, 2022 at 18:51