Question

Question #5771d

Answer 1

Here's what I got.

Explanation:

Start by taking a look at the balanced chemical equation for this combustion reaction

`"C"_3"H"_text(8(g]) + 5"O"_text(2(g]) -> 3"CO"_text(2(g]) + color(red)(4)"H"_2"O"_text((g])`

The `1:color(red)(4)` mole ratio that exists between propane, `"C"_3"H"_8`, and water tells you that the reaction produces `color(red)(4)` moles of water for every `1` mole of propane that takes part in the reaction.

Your theoretical yield will tell you what the maximum amount of water can be produced from a given amount of propane. For example, if you have one mole of propane, the theoretical yield of water will be four moles.

Simply put, the mole ratio that exists between propane and water will tell you the theoretical yield.

To determine how many moles of propane you have in that sample, use the compound's molar mass

`5 color(red)(cancel(color(black)("g"))) * ("1 mole C"_3"H"_8)/(44.01color(red)(cancel(color(black)("g")))) = "0.1136 moles C"_3"H"_8`

The reaction will produce a maximum amount of

`0.1136 color(red)(cancel(color(black)("moles C"_3"H"_8))) * (color(red)(4)" moles H"_2"O")/(1color(red)(cancel(color(black)("mole C"_3"H"_8)))) = "0.4544 moles H"_2"O"`

This is the reaction's theoretical yield, i.e. what you produce when you have a `100%` yield.

To find the mass of water produced when `"5 g"` of propane react, use water's molar mass

`0.4544 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = "8.186 g"`

You should round this off to one significant figure, but I'll leave it rounded to two sig figs, just for good measure.

The theoretical yield of water will thus be

`m_"theoretical" = color(green)("8.2 g")`

Now, you are told that the reaction has a `75%` yield. As you know, percent yield is defined as the actual yield of the reaction divided by the theoretical yield of the reaction, and multiplied by `100`.

`color(blue)("% yield" = "actual yield"/"theoretical yield" xx 100)`

You want to find the actual yield of water, so rearrange that equation to get

`"% yield" = m_"actual"/m_"theoretical" xx 100`

`m_"actual" = ("% yield" xx m_"theoretical")/100`

Plug in your value for the theoretical yield of water

`m_"actual" = (75 * "8.2 g")/100 = color(green)("6.2 g") ->` rounded to two sig figs

So, what does a `75%` yield mean?

When your reaction takes place, for every `4` molecules of water that could be produced by the reaction, only `3` molecules are actually produced, since `3` out of `4` is equivalent to `75` out of `100`, which is `75%`.