Question #732ae

1 Answer
May 29, 2017

Answer:

#71.1%#

Explanation:

The first thing you need to do here is to figure out the theoretical yield of the reaction.

Start by converting the masses of the two reactants to moles by using their respective molar masses. You will have

#42.0 color(red)(cancel(color(black)("g"))) * "1 mole TiO"_2/(79.87color(red)(cancel(color(black)("g")))) = "0.5256 moles TiO"_2#

and

#11.5 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "0.9575 moles C"#

Now, notice that the in the balanced chemical equation that describes your reaction

#"TiO"_ (2(s)) + color(red)(2)"C"_ ((s)) -> "Ti"_ ((s)) + 2"CO"_ ((g))#

every #1# mole of titanium(IV) oxide that takes part in the reaction consumes #color(red)(2)# moles of carbon.

Pick one of the two reactants and use this #1:color(red)(2)# mole ratio that exists between them to figure out if you're dealing with a limiting reagent.

I'll pick titanium(IV) oxide. In order for all the moles of titanium(IV) oxide to react, you need

#0.5256 color(red)(cancel(color(black)("moles TiO"_2))) * (color(red)(2)color(white)(.)"moles C")/(1color(red)(cancel(color(black)("mole TiO"_2)))) = "1.051 moles C"#

As you can see, you have

#overbrace("0.9575 moles C")^(color(blue)("what is available")) " "<" " overbrace("1.051 moles C")^(color(purple)("what is needed"))#

This means that the carbon will act as a limiting reagent, i.e. it will be completely consumed before all the moles of titanium(IV) oxide will get the chance to react.

Calculate the number of moles of titanium that would be produced by the reaction at #100%# yield.

#0.9575 color(red)(cancel(color(black)("moles C"))) * "1 mole Ti"/(color(red)(2)color(red)(cancel(color(black)("moles C")))) = "0.4788 moles C"#

Use the element's molar mass to convert this to moles

#0.4788 color(red)(cancel(color(black)("moles Ti"))) * "47.87 g"/(1color(red)(cancel(color(black)("mole Ti")))) = "22.92 g"#

This represents your theoretical yield, i.e. what you would expect to see at #100%# yield.

The actual yield of the reaction, which is what you get after performing it, is equal to #"16.3 g"#.

In order to find the reaction's percent yield, you must figure out the mass of titanium that is produced for every #"100 g"# of titanium that could be produced.

#100 color(red)(cancel(color(black)("g Ti"))) * "16.3 g Ti"/(22.92 color(red)(cancel(color(black)("g Ti")))) = "71.12 g Ti"#

Therefore, you can say that the reaction's percent yield is equal to

#color(darkgreen)(ul(color(black)("% yield = 71.1%")))#

The answer is rounded to three sig figs.