"50 g" of "ZnS" are strongly heated in air to effect partial oxidation and the resultant mass weighed "44 g". What is the ratio of "ZnO" to "ZnS" in the resulting mixture?

A: 13.5 : 30.5 B: 27 : 12.58 C: 27 : 15.31 D: 30.52 : 13.48

Aug 2, 2017

Here's what I got.

Explanation:

Start by taking a look at the balanced chemical equation that describes this reaction. Zinc sulfide will react with oxygen gas to produce zinc oxide and sulfur dioxide.

$2 {\text{ZnS"_ ((s)) + 3"O"_ (2(g)) -> 2"ZnO"_ ((s)) + 2"SO}}_{2 \left(g\right)}$

Now, you know that your sample of zinc sulfide undergoes partial oxidation, which means that only a fraction of the mass of zinc sulfide will react to produce zinc oxide.

Notice that zinc sulfide and zinc oxide are present in a $1 : 1$ mole ratio because it takes $2$ moles of zinc sulfide to produce $2$ moles of zinc oxide.

Use the molar masses of the two compounds to convert this mole ratio to a gram ratio.

You will have

$\text{1 mole ZnS"/"1 mole ZnO" = (1 color(red)(cancel(color(black)("mole ZnS"))) * "97.474 g"/(1 color(red)(cancel(color(black)("mole ZnS")))))/(1 color(red)(cancel(color(black)("mole ZnO"))) * "81.40 g"/(1color(red)(cancel(color(black)("mole ZnO"))))) = "97.474 g ZnS"/"81.40 g ZnO}$

So, you know that the reaction produces $\text{81.40 g}$ of zinc oxide for every $\text{97.474 g}$ of zinc sulfide that take part in the reaction.

This is equivalent to saying that when $\text{1 g}$ of zinc sulfide is oxidized, the reaction produces $\text{0.8351 g}$ of zinc oxide.

In other words, if the reaction consumes $\text{1 g}$ of zinc sulfide, the total mass of the sample will decrease by

overbrace("1 g")^(color(blue)("ZnS consumed")) - overbrace("0.8351 g")^(color(blue)("ZnO produced")) = "0.1649 g"

In your case, the total mass of the sample decreases by

$\text{50 g " - " 44 g" = "6 g}$

which means that the reaction must have consumed

6 color(red)(cancel(color(black)("g decrease"))) * "1 g ZnS"/(0.1649 color(red)(cancel(color(black)("g decrease")))) = "36.39 g"

This implies that the resultant mixture contains

overbrace("50 g ZnS")^(color(blue)("what you start with")) - overbrace("36.39 g ZnS")^(color(blue)("what is consumed")) = overbrace("13.6 g ZnS")^(color(blue)("what remains unconsumed"))

Consequently, the resultant mixture also contains

overbrace("44 g")^(color(blue)("ZnS + ZnO")) - overbrace("13.6 g")^(color(blue)("ZnS")) = overbrace("30.4 g")^(color(blue)("ZnO"))

Therefore, you can say that the resulting mixture contains zinc oxide and zinc sulfide in a $30.4 : 13.6$ gram ratio, which is pretty close to option (D).

The difference between the values was probably caused by the values I used for the molar masses of the two compounds.