Question

Question #8a08c

Answer 1

`"8.1 g"`

Explanation:

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

`"Ti"_text((s]) + color(red)(2)"F"_text(2(g]) -> "TiF"_text(4(s])`

Notice that you have `color(red)(2)` moles of fluorine gas reacting with `1` mole of titanium metal to form `1` mole of titanium(IV) fluoride.

This tells you that the reaction will always consume twice as many moles of fluorine gas as you have moles of titanium metal that are taking part in the reaction.

Now, the problem provides you with masses of reactants, so in order to make use of this `1:color(red)(2)` mole ratio that exists between them, you must convert the masses to moles.

To do that, use the molar masses of titanium metal and fluorine gas.

`5.0 color(red)(cancel(color(black)("g"))) * overbrace("1 mole Ti"/(47.867color(red)(cancel(color(black)("g")))))^(color(purple)("molar mass of Ti")) = "0.10446 moles Ti"`

`5.0color(red)(cancel(color(black)("g"))) * overbrace("1 mole F"_2/(37.997color(red)(cancel(color(black)("g")))))^(color(brown)("molar mass of F"_2)) = "0.13159 moles F"_2`

So, do you have enough moles of fluorine gas to allow for all the moles of tiatnium to react?

Use the aforementioned mole ratio to find out!

`0.10446color(red)(cancel(color(black)("moles Ti"))) * (color(red)(2)color(white)(a)"moles F"_2)/(1color(red)(cancel(color(black)("mole Ti")))) = "0.20892 moles F"_2`

That many moles of titanium would require `0.20892` moles of fluorine gas. Since you have fewer moles of fluorine gas than required, you can say that fluorine gas will act as a limiting reagent, i.e. it will determine how much titanium actually reacts.

`0.13159color(red)(cancel(color(black)("moles F"_2))) * " 1 mole Ti"/(color(red)(2)color(red)(cancel(color(black)("moles F"_2)))) = "0.06580 moles TI"`

This is how many moles of titanium will react. The rest will be in excess. The fluorine gas will be completely consumed by the reaction.

Now, the theoretical yield of the reaction tells you how many moles of product you get if the reaction has a `100%` yield.

In other words, the theoretical yield is what you get when all the moles of titanium and all the moles of fluorine gas that take part in the reaction (whatever you have in excess does not matter here) end up forming moles of product.

In this case, you have a `1:1` between titanium and titanium(IV) fluoride, so the theoretical yield of the reaction will be

`0.06580color(red)(cancel(color(black)("moles Ti"))) * "1 mole TiF"_4/(1color(red)(cancel(color(black)("mole Ti")))) = "0.06580 moles TiF"_4`

To get the theoretical yield in grams, use titanium(IV) fluoride's molar mass

`0.06580color(red)(cancel(color(black)("moles TiF"_4))) * "123.86 g"/(1color(red)(cancel(color(black)("mole TiF"_4)))) = color(green)(|bar(ul(color(white)(a/a)"8.1 g"color(white)(a/a)|)))`

The answer is rounded to two sig figs.