Question

`"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

Answer 1

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"ZnS"_ ((s)) + 3"O"_ (2(g)) -> 2"ZnO"_ ((s)) + 2"SO"_ (2(g))`

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

`"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 `"81.40 g"` of zinc oxide for every `"97.474 g"` of zinc sulfide that take part in the reaction.

This is equivalent to saying that when `"1 g"` of zinc sulfide is oxidized, the reaction produces `"0.8351 g"` of zinc oxide.

In other words, if the reaction consumes `"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

`"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.