Question

Question #3a9f2

Answer 1

`"2.10 g"`

Explanation:

Start by writing the balanced chemical equation that describes this decomposition reaction

`"ZnCO"_ (3(s)) -> "ZnO"_ ((s)) + "CO"_ (2(g)) uarr`

Zinc carbonate undergoes thermal decomposition to produce zinc oxide and carbon dioxide. As you can see, all the compounds that take part in the reaction do so in `1:1` mole ratios to each other.

Simply put, when `1` mole of zinc carbonate undergoes decomposition, `1` mole of zinc oxide and `1` mole of carbon dioxide are produced.

The problem provides you with grams of zinc carbonate, so the next thing to do is to convert the grams to moles by using the compound's molar mass.

`6.25 color(red)(cancel(color(black)("g"))) * "1 mole ZnCO"_3/(242.18color(red)(cancel(color(black)("g")))) = "0.02581 moles ZnCO"_3`

Now, the reaction produces zinc oxide in a `1:1` mole ratio to zinc carbonate, which means that you will end up with

`0.02581color(red)(cancel(color(black)("moles ZnCO"_3))) * "1 mole ZnO"/(1color(red)(cancel(color(black)("mole ZnCO"_3)))) = "0.02581 moles ZnO"`

Finally, to convert this to grams, use the compound's molar mass

`0.02581 color(red)(cancel(color(black)("moles ZnO"))) * "81.38 g"/(1color(red)(cancel(color(black)("mole ZnO")))) = color(darkgreen)(ul(color(black)("2.10 g")))`

The answer is rounded to three sig figs.

So remember, any stoichiometric problem involving a compound `"A"` and a compound `"B"` can be solved using the following approach

`color(red)(cancel(color(black)("grams of A"))) * overbrace(color(purple)(cancel(color(black)("1 mole of A")))/color(red)(cancel(color(black)("grams of A"))))^(color(black)("molar mass of A")) * overbrace(color(green)(cancel(color(black)("moles of B")))/color(purple)(cancel(color(black)("moles of A"))))^(color(black)("mole ratio between A and B")) * overbrace("grams of B"/(1color(green)(cancel(color(black)("moles of B")))))^(color(black)("molar mass of B"))`

`= " grams of B"`