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# ?

1 Answer
Aug 14, 2016

Answer:

The theoretical yield of #"TiF"_4# is 5.2 g.

Explanation:

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

We start with the balanced equation.

#color(white)(mmmmmmml)"Ti" +color(white)(l) "2F"_2 → color(white)(ll)"TiF"_4#
#"MM/g·mol"^"-1": 47.87color(white)(ll) 38.00color(white)(mll) 123.86#

(a) Identify the limiting reactant

We calculate the amount of #"TiF"_4# that can form from each reactant.

Calculate the moles of #"Ti"#.

#"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 #"TiF"_4# formed from #"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 #"F"2#

#"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 #"TiF"_4# formed from #"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 #"F"_2#, because it produces fewer moles of #"TiF"_4#.

(b) Calculate the theoretical yield of #"TiF"_4#.

#"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 #"TiF"_4# is 5.2 g.