# For this reaction Ti(s) +2F_2(g) -> TiF_4(s), what is the theoretical yield of the product, in grams, if you start with 4.8g Ti and 3.2g F_2 ?

Aug 14, 2016

The theoretical yield of ${\text{TiF}}_{4}$ is 5.2 g.

#### Explanation:

We must first identify the limiting reactant, and then we can calculate the theoretical yield.

$\textcolor{w h i t e}{m m m m m m m l} {\text{Ti" +color(white)(l) "2F"_2 → color(white)(ll)"TiF}}_{4}$
$\text{MM/g·mol"^"-1} : 47.87 \textcolor{w h i t e}{l l} 38.00 \textcolor{w h i t e}{m l l} 123.86$

(a) Identify the limiting reactant

We calculate the amount of ${\text{TiF}}_{4}$ that can form from each reactant.

Calculate the moles of $\text{Ti}$.

$\text{Moles of Ti" = 4.8 color(red)(cancel(color(black)("g Ti"))) × "1 mol Ti"/(47.87 color(red)(cancel(color(black)("g Ti")))) = "0.100 mol Ti}$

Calculate moles of ${\text{TiF}}_{4}$ formed from $\text{Ti}$

0.100color(red)(cancel(color(black)("mol Ti"))) × "1 mol TiF"_4/(1 color(red)(cancel(color(black)("mol Ti")))) = "0.100 mol TiCl"_4

Calculate the moles of $\text{F} 2$

${\text{Moles of F"_2 = 3.2 color(red)(cancel(color(black)("g F"_2))) × "1 mol F"_2/(38.00 color(red)(cancel(color(black)("g F_2")))) = "0.0842 mol F}}_{2}$

Calculate moles of ${\text{TiF}}_{4}$ formed from ${\text{F}}_{2}$

0.0842 color(red)(cancel(color(black)("mol F"_2))) × "1 mol TiF"_4/(2 color(red)(cancel(color(black)("mol F"_2)))) = "0.0421 mol TiF"_4

The limiting reactant is ${\text{F}}_{2}$, because it produces fewer moles of ${\text{TiF}}_{4}$.

(b) Calculate the theoretical yield of ${\text{TiF}}_{4}$.

${\text{Theoretical yield" = 0.0421 color(red)(cancel(color(black)("mol TiF"_4))) × "123.86 g TiF"_4/(1 color(red)(cancel(color(black)("mol TiF"_4)))) = "5.2 g TiF}}_{4}$

The theoretical yield of ${\text{TiF}}_{4}$ is 5.2 g.