Question

Question #83153

Answer 1

`75.5%`

Explanation:

Percent yield problems are all about what your reaction actually produces versus what it could theoretically produce.

`color(blue)("% yield" = "what you actually get"/"what you could theoretically get" xx 100)`

So, what your reaction actually produces is called actual yield. This represents the actual amounts of products you get at the end of the reaction.

Now, the theoretical yield of the reaction is what you calculate using the mole ratios that exist between the chemical species involved in the reaction.

The balanced chemical equation for your reaction looks like this

`"SiO"_2 + 3"C" rightleftharpoons "SiC" + 2"CO"`

Notice that you have a `1:1` mole ratio between silicon dioxide, `"SiO"_2`, and silicon carbide, `"SiC"`. This means that the reaction would theoretically produce one mole of silicon carbide for every one mole of silicon dioxide that takes part in the reaction.

You know that `991` moles of silicon dioxide react in the presence of excess carbon. This means that theoretically, the reaction should produce `991` moles of silicon carbide.

Use the molar mass of silicon carbide to determine how many moles were actually produced by the reaction - do not forget to convert the mass of silicon carbide from kilograms to grams by using the conversion factor

`"1 kg" = 10^3"g"`

`30.0 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole SiC"/(40.1 color(red)(cancel(color(black)("g")))) = "748.13 moles SiC"`

So, the reaction only produced `748.13` moles of silicon carbide, which means that its percent yield is equal to

`"% yield" = (748.13color(red)(cancel(color(black)("moles"))))/(991color(red)(cancel(color(black)("moles")))) xx 100 = color(green)("75.5%")`

The answer is rounded to three sig figs.